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**Physics 7C lecture 06 Work and Energy**

Thursday October 17, 8:00 AM – 9:20 AM Engineering Hall 1200

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Introduction The simple methods we’ve learned using Newton’s laws are inadequate when the forces are not constant. In this chapter, the introduction of the new concepts of work, energy, and the conservation of energy will allow us to deal with such problems.

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**Work A force on a body does work if the body undergoes a displacement.**

Figures 6.1 and 6.2 illustrate forces doing work.

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**Work done by a constant force**

The work done by a constant force acting at an angle to the displacement is W = Fs cos . Figure 6.3 illustrates this point. Follow Example 6.1.

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**Positive, negative, and zero work**

A force can do positive, negative, or zero work depending on the angle between the force and the displacement. Refer to Figure 6.4.

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Q6.1 An elevator is being lifted at a constant speed by a steel cable attached to an electric motor. Which statement is correct? Motor Cable v A. The cable does positive work on the elevator, and the elevator does positive work on the cable. Elevator B. The cable does positive work on the elevator, and the elevator does negative work on the cable. C. The cable does negative work on the elevator, and the elevator does positive work on the cable. D. The cable does negative work on the elevator, and the elevator does negative work on the cable. Answer: B

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A6.1 An elevator is being lifted at a constant speed by a steel cable attached to an electric motor. Which statement is correct? Motor Cable v A. The cable does positive work on the elevator, and the elevator does positive work on the cable. Elevator B. The cable does positive work on the elevator, and the elevator does negative work on the cable. C. The cable does negative work on the elevator, and the elevator does positive work on the cable. D. The cable does negative work on the elevator, and the elevator does negative work on the cable.

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Q6.2 An elevator is being lowered at a constant speed by a steel cable attached to an electric motor. Which statement is correct? Motor Cable A. The cable does positive work on the elevator, and the elevator does positive work on the cable. v Elevator B. The cable does positive work on the elevator, and the elevator does negative work on the cable. C. The cable does negative work on the elevator, and the elevator does positive work on the cable. D. The cable does negative work on the elevator, and the elevator does negative work on the cable. Answer: C

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A6.2 An elevator is being lowered at a constant speed by a steel cable attached to an electric motor. Which statement is correct? Motor Cable A. The cable does positive work on the elevator, and the elevator does positive work on the cable. v Elevator B. The cable does positive work on the elevator, and the elevator does negative work on the cable. C. The cable does negative work on the elevator, and the elevator does positive work on the cable. D. The cable does negative work on the elevator, and the elevator does negative work on the cable.

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**Work done by several forces**

W = Wtraction + Wfriction or W = (Tx + fx ) . x

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**The total work done on the sled after it has moved a distance d is**

Q6.4 A tractor driving at a constant speed pulls a sled loaded with firewood. There is friction between the sled and the road. The total work done on the sled after it has moved a distance d is A. positive. B. negative. C. zero. D. not enough information given to decide Answer: C

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A6.4 A tractor driving at a constant speed pulls a sled loaded with firewood. There is friction between the sled and the road. The total work done on the sled after it has moved a distance d is A. positive. B. negative. C. zero. D. not enough information given to decide

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**Kinetic energy The kinetic energy of a particle is K = 1/2 mv2.**

The net work on a body changes its speed and therefore its kinetic energy, as shown in Figure 6.8 below.

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**The work-energy theorem**

The work-energy theorem: The work done by the net force on a particle equals the change in the particle’s kinetic energy. Mathematically, the work-energy theorem is expressed as Wtot = K2 – K1 = K.

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**A. the object’s kinetic energy B. the object’s velocity **

Q6.5 A nonzero net force acts on an object. Which of the following quantities could be constant? A. the object’s kinetic energy B. the object’s velocity C. both of the above D. none of the above Answer: A

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A6.5 A nonzero net force acts on an object. Which of the following quantities could be constant? A. the object’s kinetic energy B. the object’s velocity C. both of the above D. none of the above

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**The work-energy theorem**

Wtot = K2 – K1 = K. Proof: F = m dV/dt and dx = V dt multiply: F dx = m V dV = d(½ m V2) = d K Since: d W = F dx we have: d W = d K or W = K2 – K1

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**Using work and energy to calculate speed**

with 1 N push, what is V2?

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**After each block has moved a distance d, the force of gravity has done**

Q6.8 Three blocks are connected as shown. The ropes and pulleys are of negligible mass. When released, block C moves downward, block B moves up the ramp, and block A moves to the right. After each block has moved a distance d, the force of gravity has done A. positive work on A, B, and C. B. zero work on A, positive work on B, and negative work on C. C. zero work on A, negative work on B, and positive work on C. D. none of these Answer: C

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A6.8 Three blocks are connected as shown. The ropes and pulleys are of negligible mass. When released, block C moves downward, block B moves up the ramp, and block A moves to the right. After each block has moved a distance d, the force of gravity has done A. positive work on A, B, and C. B. zero work on A, positive work on B, and negative work on C. C. zero work on A, negative work on B, and positive work on C. D. none of these

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**Work and energy with varying forces—Figure 6.16**

Many forces, such as the force to stretch a spring, are not constant. In Figure 6.16, we approximate the work by dividing the total displacement into many small segments.

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Stretching a spring The force required to stretch a spring a distance x is proportional to x: Fx = kx. k is the force constant (or spring constant) of the spring. The area under the graph represents the work done on the spring to stretch it a distance X: W = 1/2 kX2.

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**Motion with a varying force**

An air-track glider is attached to a spring, so the force on the glider is varying. For initial speed v1, how long can the spring extend?

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**Motion with a varying force**

For initial speed v1, how long can the spring extend? ½ m v12 = ½ k x2

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**Motion with a varying force**

For initial speed v1, how long can the spring extend? ½ m v12 = ½ k x2 + fk x

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**Motion on a curved path—Example 6.8**

A child on a swing moves along a curved path. write down all the work(s)

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**Motion on a curved path—Example 6.8**

gravity: mg R (1-cos θ) push: F R sin θ tension in the string: (Why?)

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**A. more work on the 8.00-kg block than on the 6.00-kg block. **

Q6.6 A 6.00-kg block and an 8.00-kg block are connected as shown. When released, the 6.00-kg block accelerates downward and the 8.00-kg block accelerates to the right. After each block has moved 2.00 cm, the force of gravity has done A. more work on the 8.00-kg block than on the 6.00-kg block. B. the same amount of work on both blocks. C. less work on the 8.00-kg block than on the 6.00-kg block. D. not enough information given to decide Answer: C

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A6.6 A 6.00-kg block and an 8.00-kg block are connected as shown. When released, the 6.00-kg block accelerates downward and the 8.00-kg block accelerates to the right. After each block has moved 2.00 cm, the force of gravity has done A. more work on the 8.00-kg block than on the 6.00-kg block. B. the same amount of work on both blocks. C. less work on the 8.00-kg block than on the 6.00-kg block. D. not enough information given to decide

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**A. is greater than the total work done on the 6.00-kg block. **

Q6.7 A 6.00-kg block and an 8.00-kg block are connected as shown. When released, the 6.00-kg block accelerates downward and the 8.00-kg block accelerates to the right. After each block has moved 2.00 cm, the total work done on the 8.00-kg block A. is greater than the total work done on the 6.00-kg block. B. is the same as the total work done on the 6.00-kg block. C. is less than the total work done on the 6.00-kg block. D. not enough information given to decide Answer: A

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A6.7 A 6.00-kg block and an 8.00-kg block are connected as shown. When released, the 6.00-kg block accelerates downward and the 8.00-kg block accelerates to the right. After each block has moved 2.00 cm, the total work done on the 8.00-kg block A. is greater than the total work done on the 6.00-kg block. B. is the same as the total work done on the 6.00-kg block. C. is less than the total work done on the 6.00-kg block. D. not enough information given to decide

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**Power Power is the rate at which work is done.**

Average power is Pav = W/t and instantaneous power is P = dW/dt. The SI unit of power is the watt (1 W = 1 J/s), but other familiar units are the horsepower and the kilowatt-hour.

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**A. the magnitude of the net force remains constant. **

Q6.10 An object is initially at rest. A net force (which always points in the same direction) is applied to the object so that the power of the net force is constant. As the object gains speed, A. the magnitude of the net force remains constant. B. the magnitude of the net force increases. C. the magnitude of the net force decreases. D. not enough information given to decide Answer: C

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A6.10 An object is initially at rest. A net force (which always points in the same direction) is applied to the object so that the power of the net force is constant. As the object gains speed, A. the magnitude of the net force remains constant. B. the magnitude of the net force increases. C. the magnitude of the net force decreases. D. not enough information given to decide

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**Calculate the work done by a force Kinetic energy: Ek = ½ m v2 **

Summary Calculate the work done by a force Kinetic energy: Ek = ½ m v2 work - energy theorem To relate work and kinetic energy when the forces are not constant or the body follows a curved path power P = dW/dt

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