Steve It’s “Moving Day”…. Dave tries to push a refrigerator over thick carpet. He pushes hard, but it doesn’t budge. Steve pushes an identical refrigerator across a wooden floor. He also pushes hard, and it moves easily. Who does more “work”? Steve

Θ= angle between the force and distance Steve moved the refrigerator….Dave didn’t. Work – what is accomplished when a force acts on an object and that force causes the object to move over a certain distance W = FdcosΘ W = Work (Joule) F = force (N) d = distance (m) Θ= angle between the force and distance

Is work being done by the man on the groceries by holding it in place? Is work being done by the Earth on the groceries? The net work in the vertical direction on the groceries is zero (net force = 0) Is work being done by the man on the groceries if he walks forward? No…his force on the bag is perpendicular to motion

Work is a scalar quantity -- it only has magnitude Work is a scalar quantity -- it only has magnitude. However, there is a difference between negative work and positive work. The hammer does positive work on the nail (same direction as displacement) The nail does negative work on the hammer (opposite direction of displacement)

A person pulls a 50-kg crate 40 m along a horizontal floor with a constant force of 100 N at an angle of 37 . The floor is rough and exerts a friction force of 50 N. Determine the work done by each force acting on the crate and the net work being done on the crate.

If an object’s speed changes, its kinetic energy changes If a net work is done on the bus above, what do we know about the bus It moves A force is applied It accelerates If an object’s speed changes, its kinetic energy changes Kinetic energy – energy of motion (measured in joules)

(Work-Energy Theorem) KE = 1/2 mv2 KE = kinetic energy (J) m = mass (kg) v = speed (m/s) Since work causes an object’s speed to change, it causes its kinetic energy to change (Work-Energy Theorem) ΣW = ∆KE

ΣW = 1/2 mvf2 -1/2 mvi2 If ΣW (Wnet) > 0, KE increases (object speeds up) If ΣW (Wnet) < 0, KE decreases (object slows down)

A baseball (m = 140 g) traveling 32 m/s moves a fielder’s glove backward 25 cm when the ball is caught. What was the average force exerted by the ball on the glove?

Power If you press a barbell over your head with a constant speed…. Are you doing work on the barbell? Yes Is gravity doing work on the barbell? Yes (negative) What is the net work on the barbell? Zero (no a, no ΣF) If you faster reps, you feel more fatigued. Why? Your rate of doing work is increasing.

Power = work / time P = W / t P = power (Watt) W = work (J) T = time (s) 1 Horsepower = 550 ft lb/s = 746 W

P = F d / t P = F v Since we know W = Fd, P = power (W) F = force (N) v = average speed (m/s)

A pump is to lift 18. 0 kg of water per minute to a height of 3. 60 m A pump is to lift 18.0 kg of water per minute to a height of 3.60 m. Assuming the pump moves the water at a constant speed, what power rating (watts) should the pump motor have?

A car is traveling up a 10° incline at a constant speed of 80 km/h A car is traveling up a 10° incline at a constant speed of 80 km/h. The car has a mass of 1400 kg and the frictional force of the road on the car is 700 N. How much horesepower does the motor need in order to move at that speed?

Forms of Energy Energy Mechanical Nonmechanical Kinetic Potential Gravitational Elastic Potential energy – energy associated with forces that depend on the position of an object relative to its surroundings

Gravitational Potential Energy An object gains energy when work is done to it to lift it off the ground to a position above the ground. W = Fadcosθ If object is raised at a constant speed, Fa = Fg. Therefore,

W = Fd = PEG= mgh PEg = gravitational potential energy (J) m = mass (kg) g = gravitational acceleration (m/s2) h = height (m)

A roller coaster with a mass of 1000 kg travels up a 25 m hill and then back down 10 m. What is the potential energy at the top of the hill and at the ending position on the other side?

Elastic Potential Energy A spring gains energy when work is done to it to compress or stretch it from its natural position Hooke’s Law Fp = kx Fs = -kx k = spring constant (N/m) x = amount of compression/tension (m) Fp increases as a spring is compressed, so average force is F = ½(0 + kx) = ½ kx. Therefore,

W = Fd = PES = ½kx2 PEs = elastic potential energy (J) k = spring constant (N/m) x = amount of compression/tension (m)

A spring with a force constant of 5. 2 N/m has a relaxed length of 2 A spring with a force constant of 5.2 N/m has a relaxed length of 2.45 m. When a mass is attached to the end of the spring and allowed to come to rest, the vertical length of the spring is 3.57 m. Calculate the elastic potential energy stored in the spring.

Principle of Conservation of Mechanical Energy If only conservative forces are acting , the total mechanical energy of a system neither increases nor decreases in any process. It stays constant – it is conserved. Conservative forces – work done does not depend on path; just the initial and final positions (gravitational force, elastic force) Nonconservative forces – work done depends on path taken (friction force, air resistance, applied force, tension force)

E = PE + KE E1 = E2 PE1 + KE1 = PE2 + KE2 An object’s initial E at time 1 is equal to an objects final E at time 2 assuming no nonconservative forces acting. E1 = E2 PE1 + KE1 = PE2 + KE2

If the original height of a rock is 3 If the original height of a rock is 3.0 m, calculate the rock’s speed when it has fallen 2.0 m and the speed right before it hits the ground.

A dart of mass 0.100 kg is pressed against the spring of a toy dart gun. The spring (k = 250 N/m) is compressed 6.0 cm and released. If the dart detaches from the spring when the spring reaches its natural length, what speed does the dart acquire?

Law of Conservation of Energy The total energy is neither increased nor decreased in any process. Energy can be transformed from one for to another, and transferred from one object to another, but the total amount remains constant. In a process, mechanical energy may be “lost” or “gained” due to the work that is done by nonconservative forces, but it simply is converted into or converted from another type of energy (heat, light, electric energy, etc.)

WNC = ∆KE + ∆PE The work done by nonconservative forces (to produce heat, light, etc.) is equal to the sum of the change in kinetic energy and the change in potential energy.

You drop a 2. 0-kg ball from a height of 2 You drop a 2.0-kg ball from a height of 2.0 m and it bounces back to a height of 1.5 m. How much energy is “lost” in the bounce? What is the ball’s speed at the instant it leaves the ground after the bounce?

An SUV and a sports car are both moving 60 mph down the highway. Which one has more kinetic energy? Which one has more momentum? Our intuition tells us that the SUV has more momentum because it is larger.

Is there any way to give the sports car more momentum so that it eventually is the same as the SUV? How would we make their kinetic energies equal? Increase the speed of the sports car. The same is true to make their momentums the same – increase the velocity of the sports car. Momentum - the product of an object’s mass and velocity

p = m x v p = momentum (kg x m/s) m = mass (kg) v = velocity (m/s) Momentum is different from energy in that it is a vector quantity. Direction matters! How would we change the car’s momentum? Accelerate the car by applying more force.

Σ F = ma Σ F = m(∆v/∆t) Σ F = (m∆v)/∆t Σ F = ∆p/∆t Newton’s Second Law – the net force applied to an object is equal to the rate of the change of momentum of an object

A tennis ball with a mass of 0 A tennis ball with a mass of 0.060 kg leaves a tennis racket on the serve with a velocity of 55 m/s. If the racket is in contact with the ball for 0.004 seconds, what is the force that is exerted on the ball?

Water leaves a hose at a rate of 1. 5 kg/s with a velocity of 20 m/s Water leaves a hose at a rate of 1.5 kg/s with a velocity of 20 m/s. If the water is aimed at a car window which stops it, what is the force exerted by the car on the water?

Impulse When you participate in a water balloon toss, how do you catch a water balloon? “Soft Hands” Your catch is spread over over a larger amount of time. If you were to abruptly catch the balloon, it would break.

F = ∆p/∆t F∆t = ∆p To change an object’s momentum, if the time decreases, the force has to increase (or vice versa) Impulse – the product of a force exerted on an object and the time over which the force acts

A 12-kg hammer strikes a nail at a velocity of 8 A 12-kg hammer strikes a nail at a velocity of 8.5 m/s and comes to rest in a time interval of 8.0 ms. What is the impulse given to the nail? What is the average force acting on the nail?

Law of Conservation of Momentum The total momentum of an isolated system of objects remains constant. In a system with numerous objects, the individual momentums of the objects may change as the objects interact with each other, but the sum of their momenta remains constant. mAvA0 + mBvB0 = mAvA +mBvB

Initial Momentum = Final Momentum Explosions Initially, a system has no momentum. Then, some event happens causing objects to move in various directions. Regardless of the direction or motion, all of the resultant momentums have to cancel out to equal zero (to match the initial conditions). Initial Momentum = Final Momentum mv = mAvA + mBvB

What is the recoil velocity of a 5. 0-kg rifle that shoots a 0 What is the recoil velocity of a 5.0-kg rifle that shoots a 0.020-kg bullet at a speed of 620 m/s?

Initial Momentum = Final Momentum Elastic Collisons Initially, at least one object of a system has momentum. It then hits another object causing each object to change its velocity. All of the momentums of the objects after the collision will equal the momentums of the objects before the collision. Initial Momentum = Final Momentum mAvA0 + mBvB0 = mAvA +mBvB

½mAvA02 + ½mBvB02 = ½mAvA2 + ½mBvB2 Because the collision is elastic, not only is momentum conserved, but all mechanical energy is conserved. ½mAvA02 + ½mBvB02 = ½mAvA2 + ½mBvB2 mAvA02 + mBvB02 = mAvA2 + mBvB2 mAvA02 - mAvA2 = mBvB2 - mBvB02 mA(vA02 - vA2) = mB(vB2 - vB02) mA(vA0 - vA)(vA0 + vA) = mB(vB - vB0)(vB + vB0)

mA(vA0 - vA)(vA0 + vA) = mB(vB - vB0)(vB + vB0) Back to momentum… mAvA0 + mBvB0 = mAvA +mBvB mAvA0 - mAvA = mBvB - mBvB0 mA(vA0 - vA) = mB(vB - vB0) Combining both equations… vA0 + vA = vB + vB0 vA0 - vB0 = vB – vA (for elastic collisions)

A ball of mass 0. 440 kg moving east with a speed of 2 A ball of mass 0.440 kg moving east with a speed of 2.20 m/s collides head-on with a 0.220-kg ball at rest. If the collision is perfectly elastic, what will be the velocity of each ball after the collision?

Initial Momentum = Final Momentum Inelastic Collisons Inelastic collisions are ones where two separate objects collide and stick together moving together as one unit afterward. Momentum is conserved in these types of collisions. Initial Momentum = Final Momentum mAvA + mBvB = (mA + mB)v Mechanical energy is not conserved in these types of collisions (deformation, heat, etc.)

A bullet is fired vertically into a 1 A bullet is fired vertically into a 1.40-kg block of wood at rest directly above it. If the bullet has a mass of 29.0 g and a speed of 510 m/s, how high will the block rise after the bullet becomes embedded in it?