Work and Energy Chapter 6 Lesson 1

Objectives: After completing this module, you should be able to: Define kinetic energy and potential energy, along with the appropriate units in each system. Describe the relationship between work and kinetic energy, and apply the WORK- ENERGY THEOREM. Define and apply the concept of POWER, along with the appropriate units.

Work Relates force to change in energy Scalar quantity Independent of time

Units of Work and Energy SI unit = Joule 1 J = 1 Nm = 1 kgm2/s2

Work can be positive or negative Man does positive work lifting box Man does negative work lowering box Gravity does positive work when box lowers Gravity does negative work when box is raised

Energy is the capability for doing work. Energy is anything that can be con-verted into work; i.e., anything that can exert a force through a distance. Energy is the capability for doing work.

Potential Energy Potential Energy: Ability to do work by virtue of position or condition. A suspended weight A stretched bow

Potential Energy If force depends on distance, For gravity (near Earth’s surface)

Gravitational Potential Energy Example Problem: What is the potential energy of a 50-kg person in a skyscraper if he is 480 m above the street below? Gravitational Potential Energy What is the P.E. of a 50-kg person at a height of 480 m? U = mgh = (50 kg)(9.8 m/s2)(480 m) U = 235 kJ

A speeding car or a space rocket Kinetic Energy Kinetic Energy: Ability to do work by virtue of motion. (Mass with velocity) A speeding car or a space rocket

Kinetic Energy Same units as work Remember the Eq. of motion Multiply both sides by m,

Work and Kinetic Energy A resultant force changes the velocity of an object and does work on that object. m vo vf x F

The Work-Energy Theorem Work is equal to the change in ½mv2 If we define kinetic energy as ½mv2 then we can state a very important physical principle: The Work-Energy Theorem: The work done by a resultant force is equal to the change in kinetic energy that it produces.

Examples of Kinetic Energy What is the kinetic energy of a 5-g bullet traveling at 200 m/s? 5 g 200 m/s K = 100 J What is the kinetic energy of a 1000-kg car traveling at 14.1 m/s? K = 99.4 J

Work to stop bullet = change in K.E. for bullet Example 1: A 20-g projectile strikes a mud bank, penetrating a distance of 6 cm before stopping. Find the stopping force F if the entrance velocity is 80 m/s. x F = ? 80 m/s 6 cm Work = ½ mvf2 - ½ mvo2 F x = - ½ mvo2 F (0.06 m) cos 1800 = - ½ (0.02 kg)(80 m/s)2 F = 1067 N F (0.06 m)(-1) = -64 J Work to stop bullet = change in K.E. for bullet

Example 2: A bus slams on brakes to avoid an accident Example 2: A bus slams on brakes to avoid an accident. The tread marks of the tires are 80 m long. If mk = 0.7, what was the speed before applying brakes? Work = DK Work = F(cos q) x 25 m f f = mk.n = mk mg DK = Work DK = ½ mvf2 - ½ mvo2 Work = - mk mg x -½ mvo2 = -mk mg x vo = 2mkgx vo = 59.9 m/s vo = 2(0.7)(9.8 m/s2)(25 m)

Example 3: A 4-kg block slides from rest at top to bottom of the 300 inclined plane. Find velocity at bottom. (h = 20 m and mk = 0.2) Plan: We must calculate both the resultant work and the net displacement x. Then the velocity can be found from the fact that Work = DK. h 300 n f mg x Resultant work = (Resultant force down the plane) x (the displacement down the plane)

From trig, we know that the Sin 300 = h/x and: Example 3 (Cont.): We first find the net displacement x down the plane: h 300 n f mg x h x 300 From trig, we know that the Sin 300 = h/x and:

Draw free-body diagram to find the resultant force: Example 3(Cont.): Next we find the resultant work on 4-kg block. (x = 40 m and mk = 0.2) h 300 n f mg x = 40 m Draw free-body diagram to find the resultant force: x y mg cos 300 mg sin 300 Wx = (4 kg)(9.8 m/s2)(sin 300) = 19.6 N Wy = (4 kg)(9.8 m/s2)(cos 300) = 33.9 N

Example 3(Cont. ): Find the resultant force on 4-kg block Example 3(Cont.): Find the resultant force on 4-kg block. (x = 40 m and mk = 0.2) Resultant force down plane: 19.6 N - f n f mg 300 x y 33.9 N 19.6 N Recall that fk = mk n SFy = 0 or n = 33.9 N Resultant Force = 19.6 N – mkn ; and mk = 0.2 Resultant Force = 19.6 N – (0.2)(33.9 N) = 12.8 N Resultant Force Down Plane = 12.8 N

Example 3 (Cont. ): The resultant work on 4-kg block Example 3 (Cont.): The resultant work on 4-kg block. (x = 40 m and FR = 12.8 N) (Work)R = FRx x Net Work = (12.8 N)(40 m) FR 300 Net Work = 512 J Finally, we are able to apply the work-energy theorem to find the final velocity:

Work done on block equals the change in K. E. of block. Example 3 (Cont.): A 4-kg block slides from rest at top to bottom of the 300 plane. Find velocity at bottom. (h = 20 m and mk = 0.2) h 300 n f mg x Resultant Work = 512 J Work done on block equals the change in K. E. of block. ½ mvf2 - ½ mvo2 = Work ½ mvf2 = 512 J ½(4 kg)vf2 = 512 J vf = 16 m/s

Three things are necessary for the performance of work: q x There must be an applied force F. There must be a displacement x. The force must have a component along the displacement.

If a force does not affect displacement, it does no work. The force F exerted on the pot by the man does work. F W The earth exerts a force W on pot, but does no work even though there is displacement.

Work = Force component X displacement Definition of Work Work is a scalar quantity equal to the product of the displacement x and the component of the force Fx in the direction of the displacement. Work = Force component X displacement Work = Fx x

Positive Work F x Force F contributes to displacement x. Example: If F = 40 N and x = 4 m, then Work = (40 N)(4 m) = 160 Nm Work = 160 J 1 Nm = 1 Joule (J)

Negative Work x f The friction force f opposes the displacement. Example: If f = -10 N and x = 4 m, then Work = (-10 N)(4 m) = - 40 J Work = - 40 J

Resultant Work or Net Work Resultant work is the algebraic sum of the individual works of each force. F x f Example: F = 40 N, f = -10 N and x = 4 m Work = (40 N)(4 m) + (-10 N)(4 m) Work = 120 J

Resultant work is also equal to the work of the RESULTANT force. Resultant Work (Cont.) Resultant work is also equal to the work of the RESULTANT force. 40 N 4 m -10 N Example: Work = (F - f) x Work = (40 - 10 N)(4 m) Work = 120 J

Work of a Force at an Angle x = 12 m F = 70 N 60o Work = Fx x Work = (F cos ) x Work = (70 N) Cos 600 (12 m) = 420 J Only the x-component of the force does work! Work = 420 J

Procedure for Calculating Work 1. Draw sketch and establish what is given and what is to be found. Procedure for Calculating Work 2. Draw free-body diagram choosing positive x-axis along displacement. Work = (F cos ) x + F q x n mg x 3. Find work of a single force from formula. 4. Resultant work is work of resultant force.

Note: Work is positive since Fx and x are in the same direction. Example 1: A lawn mower is pushed a horizontal distance of 20 m by a force of 200 N directed at an angle of 300 with the ground. What is the work of this force? 300 x = 20 m F = 200 N F Work = (F cos q ) x Note: Work is positive since Fx and x are in the same direction. Work = (200 N)(20 m) Cos 300 Work = 3460 J

n x P 1. Draw sketch and find given values. +x x mg 8 m P fk Example 2: A 40-N force pulls a 4-kg block a horizontal distance of 8 m. The rope makes an angle of 350 with the floor and uk = 0.2. What is the work done by each acting on block? x P q 1. Draw sketch and find given values. P = 40 N; x = 8 m, uk = 0.2; q = 350; m = 4 kg +x 40 N 350 x n mg 8 m P fk 2. Draw free-body diagram showing all forces. (Cont.) Work = (F cos ) x

Example 2 (Cont.): Find Work Done by Each Force. 350 x n W = mg 8 m P fk P = 40 N; x = 8 m, uk = 0.2; q = 350; m = 4 kg 4. First find work of P. Work = (P cos ) x WorkP = (40 N) cos 350 (8 m) = 262 J 5. Next consider normal force n and weight W. Each makes a 900 angle with x, so that the works are zero. (cos 900=0): WorkP = 0 Workn = 0

6. Next find work of friction. Example 2 (Cont.): +x 40 N 350 x n W = mg 8 m P fk P = 40 N; x = 8 m, uk = 0.2; q = 350; m = 4 kg WorkP = 262 J Workn = WorkW = 0 Recall: fk = mk n 6. Next find work of friction. n + P sin 350 – mg = 0; n = mg – P cos 350 n = (4 kg)(9.8 m/s2) – (40 N)sin 350 = 16.3 N fk = mk n = (0.2)(16.3 N); fk = 3.25 N

n +x x 8 m P fk WorkP = 262 J Workn = WorkW = 0 Example 2 (Cont.): +x 40 N 350 x n W = mg 8 m P fk WorkP = 262 J Workn = WorkW = 0 6. Work of friction (Cont.) fk = 3.25 N; x = 8 m Workf = (3.25 N) cos 1800 (8 m) = -26.0 J Note work of friction is negative cos 1800 = -1 7. The resultant work is the sum of all works: 262 J + 0 + 0 – 26 J (Work)R = 236 J

To Be Continued…