Work

Work is the product of the magnitude of the __________________ moved times the component of a ________________ in the direction of the ________________.

Work Work is the product of the magnitude of the displacement moved times the component of a force in the direction of the displacement.

Work Work is the product of the magnitude of the displacement moved times the component of a force in the direction of the displacement. F Δd | F | cosθ θ

Work Work is the product of the magnitude of the displacement moved times the component of a force in the direction of the displacement. Defining equation: ? F Δd | F | cosθ θ

Work Work is the product of the magnitude of the displacement moved times the component of a force in the direction of the displacement. Defining equation: W = | Δd | X | F | cosθ where θ is the smallest angle between Δd and F in a “ tail-to-tail “ diagram F Δd | F | cosθ θ

Work W = | Δd | X | F | cosθ W can be written using a special mathematical “vector operator” How?

Work W = | Δd | X | F | cosθ W can be written using a special mathematical “vector operator” W = Δd ● F

Work W = | Δd | X | F | cosθ W can be written using a special mathematical “vector operator” W = Δd ● F note that ● stands for “dot” or “scalar” product

Work W = | Δd | X | F | cosθ W can be written using a special mathematical “vector operator” W = Δd ● F note that ● stands for “dot” or “scalar” product So Δd ● F = | Δd | X | F | cosθ

Work W = | Δd | X | F | cosθ W can be written using a special mathematical “vector operator” W = Δd ● F note that ● stands for “dot” or “scalar” product So Δd ● F = | Δd | X | F | cosθ Since Work is a product of magnitudes only, it is a _____ quantity.

Work W = | Δd | X | F | cosθ W can be written using a special mathematical “vector operator” W = Δd ● F note that ● stands for “dot” or “scalar” product So Δd ● F = | Δd | X | F | cosθ Since Work is a product of magnitudes only, it is a scalar quantity.

Work So Work can be defined mathematically as... W = Δd ● F or | Δd | X | F | cosθ

Work So Work can be defined mathematically as... W = Δd ● F or | Δd | X | F | cosθ What is the SI unit of work?

Work So Work can be defined mathematically as... W = Δd ● F or | Δd | X | F | cosθ SI unit is the Joule = ?

Work So Work can be defined mathematically as... W = Δd ● F or | Δd | X | F | cosθ SI unit is the Joule = N m = ?

Work So Work can be defined mathematically as... W = Δd ● F or | Δd | X | F | cosθ SI unit is the Joule = N m = ( kg m /s 2 ) m = ?

Work So Work can be defined mathematically as... W = Δd ● F or | Δd | X | F | cosθ SI unit is the Joule = N m = ( kg m /s 2 ) m = kg m 2 / s 2

Work So Work can be defined mathematically as... W = Δd ● F or | Δd | X | F | cosθ SI unit is the Joule = N m = ( kg m /s 2 ) m = kg m 2 / s 2 Note that “F” in the formula can be any kind of force

Work So Work can be defined mathematically as... W = Δd ● F or | Δd | X | F | cosθ SI unit is the Joule = N m = ( kg m /s 2 ) m = kg m 2 / s 2 Note that “F” in the formula can be any kind of force Both the force doing the work and the object that the work is being done on must be specified or understood.

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N a) What is the work done on the box by the applied force?

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N a) What is the work done on the box by the applied force? Draw a tail-to-tail diagram F a = 100 N θ = 36.9° Δd = 5.00 m F a = 100 N θ = 36.9° Δd = 5.00 m F a = 100 N θ = 36.9° Δd = 5.00 m

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N a) What is the work done on the box by the applied force? W = | Δd | X | F | cosθ F a = 100 N θ = 36.9° Δd = 5.00 m F a = 100 N θ = 36.9° Δd = 5.00 m F a = 100 N θ = 36.9° Δd = 5.00 m

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N a) What is the work done on the box by the applied force? W = | Δd | X | F | cosθ = 5.00 m X 100.0N cos 36.9° F a = 100 N θ = 36.9° Δd = 5.00 m F a = 100 N θ = 36.9° Δd = 5.00 m F a = 100 N θ = 36.9° Δd = 5.00 m

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N a) What is the work done on the box by the applied force? W = | Δd | X | F | cosθ = 5.00 m X 100.0N cos 36.9° = 400 J F a = 100 N θ = 36.9° Δd = 5.00 m F a = 100 N θ = 36.9° Δd = 5.00 m F a = 100 N θ = 36.9° Δd = 5.00 m

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N b) What is the work done on the box by the net force?

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N b) What is the work done on the box by the net force? Find the net force. It acts along the horizontal-axis.

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N b) What is the work done on the box by the net force? F net = 100 cos 36.9° - 22 = ?

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N b) What is the work done on the box by the net force? F net = 100 cos 36.9° - 22 = 58.0 N [E]

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N b) What is the work done on the box by the net force? Draw a tail-to-tail diagram

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N b) What is the work done on the box by the net force? Draw a tail-to-tail diagram F net = 58.0 N Δd = 5.00 m

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N b) What is the work done on the box by the net force? F net = 58.0 N Δd = 5.00 m W = | Δd | X | F | cosθ

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N b) What is the work done on the box by the net force? F net = 58.0 N Δd = 5.00 m W = | Δd | X | F | cosθ = 5.00 m X 58.0 N cos 0°

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N b) What is the work done on the box by the net force? F net = 58.0 N Δd = 5.00 m W = | Δd | X | F | cosθ = 5.00 m X 58.0 N cos 0° = 290 J

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N b) What is the work done on the box by the net force? F net = 58.0 N Δd = 5.00 m W = | Δd | X | F | cosθ = 5.00 m X 58.0 N cos 0° = 290 J Note that if θ = 0, then a short-cut formula can be used W = F d

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N b) What is the work done on the box by the net force? F net = 58.0 N Δd = 5.00 m W = | Δd | X | F | cosθ = 5.00 m X 58.0 N cos 0° = 290 J Note that if θ = 0, then a short-cut formula can be used W = F d The work done by the net force is sometimes called the “net work done” or W net

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N c) What is the work done on the box by the kinetic friction force?

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N c) What is the work done on the box by the kinetic friction force? Draw a tail-to-tail diagram

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N c) What is the work done on the box by the kinetic friction force? Draw a tail-to-tail diagram f K = 22.0 N Δd = 5.00 m θ = ?

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N c) What is the work done on the box by the kinetic friction force? Draw a tail-to-tail diagram f K = 22.0 N Δd = 5.00 m θ = 180°

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N c) What is the work done on the box by the kinetic friction force? f K = 22.0 N Δd = 5.00 m θ = 180° W = | Δd | X | F | cosθ = ?

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N c) What is the work done on the box by the kinetic friction force? f K = 22.0 N Δd = 5.00 m θ = 180° W = | Δd | X | F | cosθ =5.00 m X 22 N cos 180°

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N c) What is the work done on the box by the kinetic friction force? f K = 22.0 N Δd = 5.00 m θ = 180° W = | Δd | X | F | cosθ =5.00 m X 22 N cos 180° = J

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N c) What is the work done on the box by the kinetic friction force? f K = 22.0 N Δd = 5.00 m θ = 180° W = | Δd | X | F | cosθ =5.00 m X 22 N cos 180° = J The negative sign means energy is lost to the surroundings as heat.

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N c) What is the work done on the box by the kinetic friction force? f K = 22.0 N Δd = 5.00 m θ = 180° W = | Δd | X | F | cosθ =5.00 m X 22 N cos 180° = J The negative sign means energy is lost to the surroundings as heat. Note that f k has θ = 180°, so a short-cut formula can be used W = - f k d

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N d) What is the work done on the box by the weight?

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N d) What is the work done on the box by the weight? Draw a tail-to-tail diagram

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N d) What is the work done on the box by the weight? Draw a tail-to-tail diagram F g = 480 N Δd = 5.00 m

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N d) What is the work done on the box by the weight? Draw a tail-to-tail diagram F g = 480 N Δd = 5.00 m θ = ?

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N d) What is the work done on the box by the weight? Draw a tail-to-tail diagram F g = 480 N Δd = 5.00 m θ = 90°

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N d) What is the work done on the box by the weight? F g = 480 N Δd = 5.00 m θ = 90° W = | Δd | X | F | cosθ = ?

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N d) What is the work done on the box by the weight? F g = 480 N Δd = 5.00 m θ = 90° W = | Δd | X | F | cosθ = 5.00 m X 480N cos 90° = ?

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N d) What is the work done on the box by the weight? F g = 480 N Δd = 5.00 m θ = 90° W = | Δd | X | F | cosθ = 5.00 m X 480N cos 90° = 0 J

Example: A worker pulls on 48.0 kg box using a rope as shown. He drags the box a displacement of 5.00 m [E] horizontally along the rough floor with an applied force of N at an angle of 36.9° above the horizontal. The kinetic friction force acting is 22.0 N. Rough floor m = 48.0 kg F a = 100 N θ = 36.9° Δd = 5.00 m f K = 22.0 N d) What is the work done on the box by the weight? F g = 480 N Δd = 5.00 m θ = 90° W = | Δd | X | F | cosθ = 5.00 m X 480N cos 90° = 0 J Note that if θ = 90°, no work is done

Conditions where no work is done

1. If θ = ?

Conditions where no work is done 1. If θ = 90°

Conditions where no work is done 1. If θ = 90° 2. If | F | = ?

Conditions where no work is done 1. If θ = 90° 2. If | F | = 0 N

Conditions where no work is done 1. If θ = 90° 2. If | F | = 0 N 3. If | Δd | = ?

Conditions where no work is done 1. If θ = 90° 2. If | F | = 0 N 3. If | Δd | = 0 m

Energy

Definition in words: ?

Energy Energy is the ability to do work

Energy Energy is the ability to do work Since energy is defined in terms of “work”, it is a __________ quantity.

Energy Energy is the ability to do work Since energy is defined in terms of “work”, it is a scalar quantity.

Energy Energy is the ability to do work Since energy is defined in terms of “work”, it is a scalar quantity. Like work, the SI unit for any type of energy is the __________.

Energy Energy is the ability to do work Since energy is defined in terms of “work”, it is a scalar quantity. Like work, the SI unit for any type of energy is the Joule.

Energy of Motion

Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies.

Energy of Motion Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies. Energy that an object possesses by virtue of its motion is called ____________ energy.

Energy of Motion Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies. Energy that an object possesses by virtue of its motion is called kinetic energy.

Energy of Motion Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies. Energy that an object possesses by virtue of its motion is called kinetic energy. The symbol for kinetic energy is ______.

Energy of Motion Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies. Energy that an object possesses by virtue of its motion is called kinetic energy. The symbol for kinetic energy is E k.

Energy of Motion Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies. Energy that an object possesses by virtue of its motion is called kinetic energy. The symbol for kinetic energy is E k. What is the formula for kinetic energy?

Energy of Motion Moving objects can collide with other bodies and exert a “force through a distance” on these bodies, so moving objects are able to do work on other bodies. Energy that an object possesses by virtue of its motion is called kinetic energy. The symbol for kinetic energy is E k. E k = mv 2 /2

Try this example: Compare the kinetic energy of a 5055 kg truck and a 20 g bee moving at 36.0 km/h. E k = mv 2 /2

Truck m = 5055 kg v = 36.0 km/h = 10.0 m/s E k = ? E k = mv 2 /2 = 5055(10.0) 2 /2 = 2.5 X 10 5 J Bee m = kg v = 36.0 km/h = 10.0 m/s E k = ? E k = mv 2 /2 = 0.020(10.0) 2 /2 = 1.0 J

Try this example #2 : An 8.00 kg bird has a kinetic energy of 576 J. How fast is it moving?

Given: m = 8.00 kg E k = 576 J Unknown: v = ? Formula: E k = mv 2 /2 so v = (2 E k / m ) 1/2 Sub: v = (2 X 576 / 8.00 ) 1/2 = 12 m/s

Energy of raised position

An object that is raised above a certain reference level “is able” or has “potential” to do work.

Energy of raised position An object that is raised above a certain reference level “is able” or has “potential” to do work. If the object falls, the force of earth's gravity can exert a “force through a distance” on that object and do work on it.

Energy of raised position An object that is raised above a certain reference level “is able” or has “potential” to do work. If the object falls, the force of earth's gravity can exert a “force through a distance” on that object and do work on it. The energy of raised position is called _________ __________ energy.

Energy of raised position An object that is raised above a certain reference level “is able” or has “potential” to do work. If the object falls, the force of earth's gravity can exert a “force through a distance” on that object and do work on it. The energy of raised position is called gravitational potential energy. Its symbol is ____.

Energy of raised position An object that is raised above a certain reference level “is able” or has “potential” to do work. If the object falls, the force of earth's gravity can exert a “force through a distance” on that object and do work on it. The energy of raised position is called gravitational potential energy. Its symbol is E g.

Energy of raised position An object that is raised above a certain reference level “is able” or has “potential” to do work. If the object falls, the force of earth's gravity can exert a “force through a distance” on that object and do work on it. The energy of raised position is called gravitational potential energy. Its symbol is E g. The formula for gravitational potential energy is ___________

Energy of raised position An object that is raised above a certain reference level “is able” or has “potential” to do work. If the object falls, the force of earth's gravity can exert a “force through a distance” on that object and do work on it. The energy of raised position is called gravitational potential energy. Its symbol is E g. The formula for gravitational potential energy is E g = mgh

More About the formula for gravitational potential energy

E g = mgh

More About the formula for gravitational potential energy E g = mgh Note that the height or “h” and therefore E g depends on the reference level used to specify E g = 0 or the zero level of gravitational potential energy.

More About the formula for gravitational potential energy E g = mgh Note that the height or “h” and therefore E g depends on the reference level used to specify E g = 0 or the zero level of gravitational potential energy. Usually “ground level” is the specified zero reference level but not always. The zero reference level could be the top of a table or the bottom of the swing of a pendulum.

More About the formula for gravitational potential energy E g = mgh Note that the height or “h” and therefore E g depends on the reference level used to specify E g = 0 or the zero level of gravitational potential energy. Usually “ground level” is the specified zero reference level but not always. The zero reference level could be the top of a table or the bottom of the swing of a pendulum. Note “h” can be + or – depending on whether it is “above' or “below” the zero reference level.

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m a) E g = mgh = ?

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m a) E g = mgh = 14 X 10.0 X 6.0

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m a) E g = mgh = 14 X 10.0 X 6.0 = ?

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m a) E g = mgh = 14 X 10.0 X 6.0 = 840 J

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m a) E g = mgh = 14 X 10.0 X 6.0 = 840 J b) E g = mgh = ?

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m a) E g = mgh = 14 X 10.0 X 6.0 = 840 J b) E g = mgh = 14 X 10.0 ( -4 )

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m a) E g = mgh = 14 X 10.0 X 6.0 = 840 J b) E g = mgh = 14 X 10.0 ( -4 ) = ?

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m a) E g = mgh = 14 X 10.0 X 6.0 = 840 J b) E g = mgh = 14 X 10.0 ( -4 ) = J

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m a) E g = mgh = 14 X 10.0 X 6.0 = 840 J b) E g = mgh = 14 X 10.0 ( -4 ) = J c) E g = mgh = ?

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m a) E g = mgh = 14 X 10.0 X 6.0 = 840 J b) E g = mgh = 14 X 10.0 ( -4 ) = J c) E g = mgh = 14 X 10.0 (0)

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m a) E g = mgh = 14 X 10.0 X 6.0 = 840 J b) E g = mgh = 14 X 10.0 ( -4 ) = J c) E g = mgh = 14 X 10.0 (0) = ?

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m a) E g = mgh = 14 X 10.0 X 6.0 = 840 J b) E g = mgh = 14 X 10.0 ( -4 ) = J c) E g = mgh = 14 X 10.0 (0) = 0 J

Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge m = 14 kg ground ledge Cliff top 4.0 m 6.0 m a) E g = mgh = 14 X 10.0 X 6.0 = 840 J b) E g = mgh = 14 X 10.0 ( -4 ) = J c) E g = mgh = 14 X 10.0 (0) = 0 J Temperature is another scalar quantity like E g and it can have negative values too. It just depends on the zero reference level as well. The freezing point of water is 0° C in the Celsius scale but 273° K in the Kelvin scale.

Energy stored in a stretched or compressed spring

Like a slingshot, a stretched or compressed spring can exert a “force through a distance” and “is able to do work” on an object. Therefore, a deformed spring has energy.

Energy stored in a stretched or compressed spring Like a slingshot, a stretched or compressed spring can exert a “force through a distance” and “is able to do work” on an object. Therefore, a deformed spring has energy. The energy of a deformed spring is called ___________ ___________ energy.

Energy stored in a stretched or compressed spring Like a slingshot, a stretched or compressed spring can exert a “force through a distance” and “is able to do work” on an object. Therefore, a deformed spring has energy. The energy of a deformed spring is called elastic potential energy.

Energy stored in a stretched or compressed spring Like a slingshot, a stretched or compressed spring can exert a “force through a distance” and “is able to do work” on an object. Therefore, a deformed spring has energy. The energy of a deformed spring is called elastic potential energy. The symbol for elastic potential energy is _____.

Energy stored in a stretched or compressed spring Like a slingshot, a stretched or compressed spring can exert a “force through a distance” and “is able to do work” on an object. Therefore, a deformed spring has energy. The energy of a deformed spring is called elastic potential energy. The symbol for elastic potential energy is E e.

Energy stored in a stretched or compressed spring Like a slingshot, a stretched or compressed spring can exert a “force through a distance” and “is able to do work” on an object. Therefore, a deformed spring has energy. The energy of a deformed spring is called elastic potential energy. The symbol for elastic potential energy is E e. To understand further, we need to know about forces on deformed springs.

Forces needed to “deform” a spring

Equilibrium or rest position stretched X X compression FsFs FsFs

Forces needed to “deform” a spring Equilibrium or rest position stretched X X compression FsFs FsFs X = amount of compression or stretch of a spring from its rest position (deformation)

Forces needed to “deform” a spring Equilibrium or rest position stretched X X compression FsFs FsFs X = amount of compression or stretch of a spring from its rest position (deformation) F s = magnitude of force needed to deform a spring

Forces needed to “deform” a spring Equilibrium or rest position stretched X X compression FsFs FsFs X = amount of compression or stretch of a spring from its rest position (deformation) F s = magnitude of force needed to deform a spring Robert Hooke discovered that F s α X Hooke's law

More on Hooke's Law

F s α X

More on Hooke's Law F s α X How can we change this to an equation?

More on Hooke's Law F s α X F s = KX

More on Hooke's Law F s α X F s = KX K, the constant of proportionality, is called the __________ constant of the spring or __________ constant of the spring or just the _________ constant.

More on Hooke's Law F s α X F s = KX K, the constant of proportionality, is called the force constant of the spring or __________ constant of the spring or just the _________ constant.

More on Hooke's Law F s α X F s = KX K, the constant of proportionality, is called the force constant of the spring or Hooke's constant of the spring or just the _________ constant.

More on Hooke's Law F s α X F s = KX K, the constant of proportionality, is called the force constant of the spring or Hooke's constant of the spring or just the spring constant.

More on Hooke's Law F s α X F s = KX K, the constant of proportionality, is called the force constant of the spring or Hooke's constant of the spring or just the spring constant. What does K depend on?

More on Hooke's Law F s α X F s = KX K, the constant of proportionality, is called the force constant of the spring or Hooke's constant of the spring or just the spring constant. K depends on the “stiffness” of the spring involved

More on Hooke's Law F s α X F s = KX K, the constant of proportionality, is called the force constant of the spring or Hooke's constant of the spring or just the spring constant. K depends on the “stiffness” of the spring involved If F s = KX, then K = ?

More on Hooke's Law F s α X F s = KX K, the constant of proportionality, is called the force constant of the spring or Hooke's constant of the spring or just the spring constant. K depends on the “stiffness” of the spring involved If F s = KX, then K = F s / X

More on Hooke's Law F s α X F s = KX K, the constant of proportionality, is called the force constant of the spring or Hooke's constant of the spring or just the spring constant. K depends on the “stiffness” of the spring involved If F s = KX, then K = F s / X and SI unit for K is ______.

More on Hooke's Law F s α X F s = KX K, the constant of proportionality, is called the force constant of the spring or Hooke's constant of the spring or just the spring constant. K depends on the “stiffness” of the spring involved If F s = KX, then K = F s / X and SI unit for K is N/m.

Hooke's Law Example: A kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring. X = 8.00 cm m =0.500 kg rest

Hooke's Law Example: A kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring. X = 8.00 cm m =0.500 kg rest Draw an FBD of the mass

Hooke's Law Example: A kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring. X = 8.00 cm m =0.500 kg rest FsFs FgFg

Hooke's Law Example: A kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring. X = 8.00 cm m =0.500 kg rest F nety = ? FsFs FgFg

Hooke's Law Example: A kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring. X = 8.00 cm m =0.500 kg rest F nety = 0 FsFs FgFg

Hooke's Law Example: A kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring. X = 8.00 cm m =0.500 kg rest F nety = 0 F s + F g = 0 FsFs FgFg

Hooke's Law Example: A kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring. X = 8.00 cm m =0.500 kg rest F nety = 0 F s + F g = 0 +KX – 5.00 = 0 FsFs FgFg

Hooke's Law Example: A kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring. X = 8.00 cm m =0.500 kg rest F nety = 0 F s + F g = 0 +KX – 5.00 = 0 +K(.0800) – 5.00 = 0 FsFs FgFg

Hooke's Law Example: A kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring. X = 8.00 cm m =0.500 kg rest F nety = 0 F s + F g = 0 +KX – 5.00 = 0 +K(.0800) – 5.00 = 0 K(.0800) = 5.00 FsFs FgFg

Hooke's Law Example: A kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring. X = 8.00 cm m =0.500 kg rest F nety = 0 F s + F g = 0 +KX – 5.00 = 0 +K(.0800) – 5.00 = 0 K(.0800) = 5.00 K= 5.00 /.0800 FsFs FgFg

Hooke's Law Example: A kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring. X = 8.00 cm m =0.500 kg rest F nety = 0 F s + F g = 0 +KX – 5.00 = 0 +K(.0800) – 5.00 = 0 K(.0800) = 5.00 K= 5.00 /.0800 K= 62.5 N/m FsFs FgFg

Hooke's Law Example: A kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring. X = 8.00 cm m =0.500 kg rest F nety = 0 F s + F g = 0 +KX – 5.00 = 0 +K(.0800) – 5.00 = 0 K(.0800) = 5.00 K= 5.00 /.0800 K= 62.5 N/m FsFs FgFg This k value tells us that this spring requires 62.5 N of force to stretch or compress it by 1 meter. However, most strings have a threshold limit of deformation, beyond which Hooke's law does not hold.

Elastic potential energy

Let's plot F s vs X FsFs X

Elastic potential energy Let's plot F s vs X FsFs X

Elastic potential energy Let's plot F s vs X FsFs X The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve.

Elastic potential energy Let's plot F s vs X FsFs X The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the spring is stretched X meters as shown

Elastic potential energy Let's plot F s vs X FsFs X The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the spring is stretched X meters as shown In terms of X, what is F s = ?

Elastic potential energy Let's plot F s vs X FsFs X The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the spring is stretched X meters as shown In terms of X, F s = KX

Elastic potential energy Let's plot F s vs X FsFs X The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the spring is stretched X meters as shown In terms of X, F s = KX Elastic potential energy, E e = the area under the curve. The area is a triangle. What is the formula for the area of a triangle?

Elastic potential energy Let's plot F s vs X FsFs X The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the spring is stretched X meters as shown In terms of X, F s = KX Elastic potential energy, E e = the area under the curve. The area is a triangle. A = b h / 2

Elastic potential energy Let's plot F s vs X FsFs X The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the spring is stretched X meters as shown In terms of X, F s = KX Elastic potential energy, E e = the area under the curve. The area is a triangle. A = b h / 2 Therefore E e = ?

Elastic potential energy Let's plot F s vs X FsFs X The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the spring is stretched X meters as shown In terms of X, F s = KX Elastic potential energy, E e = the area under the curve. The area is a triangle. A = b h / 2 Therefore E e = (X ) ( KX ) / 2

Elastic potential energy Let's plot F s vs X FsFs X The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the spring is stretched X meters as shown In terms of X, F s = KX Elastic potential energy, E e = the area under the curve. The area is a triangle. A = b h / 2 Therefore E e = (X ) ( KX ) / 2 E e = KX 2 / 2

Elastic potential energy Let's plot F s vs X FsFs X The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Let's assume the spring is stretched X meters as shown In terms of X, F s = KX Elastic potential energy, E e = the area under the curve. The area is a triangle. A = b h / 2 Therefore E e = (X ) ( KX ) / 2 E e = KX 2 / 2 The elastic potential energy formula is... E e = KX 2 / 2

Try this! The spring in the previous example has a force constant of 62.5 N/m and is stretched 8.00 cm. How much elastic potential energy does this stretched spring have?

Given: K = 62.5 N/m X = 8.00 cm or m Unknown: E e = ? Formula: E e = KX 2 / 2 Sub: E e = (62.5)(0.0800) 2 / 2 = J Answer: The spring has J of elastic potential energy. It is able to do J of work.