5.1-5.2 Worksheet: Work (W) and Kinetic Energy (KE)

IB Physics SLGOHS

5.1-5.2 Worksheet: Work (W) and Kinetic Energy (KE)

Conceptual Questions

2. Discuss whether any work is being done by each of the following agents and, if so, whether the work is positive ornegative: (a) a chicken scratching the ground, (b) a personstudying, (c) a crane lifting a bucket of concrete,(d) the force of gravity on the bucket in part (c), (e) theleg muscles of a person in the act of sitting down.

4. (a) Can the kinetic energy of a system be negative? (b) Can the gravitational potential energy of a system be negative? Explain.

6. (a) If the speed of a particle is doubled, what happens to its kinetic energy? (b) If the net work done on a particle is zero, what can be said about its speed?

Problems

1. A weight lifter lifts a 350-N set of weights from ground level to a position over his head, a vertical distance of

2.00 m. How much work does the weight lifter do, assuming he moves the weights at constant speed?

2. If a man lifts a 20.0-kg bucket from a well and does 6.00 kJ of work, how deep is the well? Assume that the speed of the bucket remains constant as it is lifted.

5. Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0° incline. The coefficient of kinetic friction between the block and the incline is k _ 0.436. Determine (a) the work done by the force of gravity, (b) the net Work if the force of friction is 18.5 N and (c) the work done by the normal force.

8.A block of mass 2.50 kg is pushed 2.20 m along a frictionless horizontal table by a constant 16.0-N force directed 25.0° below the horizontal. Determine the work done by (a) the applied force, (b) the normal force exerted by thetable, (c) the force of gravity, and (d) the net force on the block.

MORE ON BACK!!!!!

9. A mechanic pushes a 2.50 _ 103-kg car from rest to a speed of v, doing 5 000 J of work in the process. Duringthis time, the car moves 25.0 m. Neglecting friction between car and road, find (a) v and (b) the horizontal force exerted on the car.

10. A 7.00-kg bowling ball moves at 3.00 m/s. How fast must a 2.45-g Ping-Pong ball move so that the two balls have the same kinetic energy?

11. A person doing a chin-up weighs 700 N, exclusive of the arms. During the first 25.0 cm of the lift, each arm exerts an upward force of 355 N on the torso. If the upward movement starts from rest, what is the person’s velocity at that point?

15.A 2.0-g bullet leaves the barrel of a gun at a speed of 300 m/s. (a) Find its kinetic energy. (b) Find the average force exerted by the expanding gases on the bullet as it moves the length of the 50-cm-long barrel.

ANSWERS:

Concept Questions

2.(a) The chicken does positive work on the ground. (b) No work is done. (c) The crane does positive work on the bucket. (d) The force of gravity does negative work on the bucket. (e)The leg muscles do negative work on the individual.

4.(a) Kinetic energy is always positive. Mass and speed squared are both positive. (b)Gravitational potential energy can be negative when the object is lower than the chosen reference level.

6.(a) Kinetic energy is proportional to the speed squared. Doubling the speed makes the object’s kinetic energy four times larger. (b) If the total work done on an object in some process is zero, its speed must be the same at the final point as it was at the initial point.

Problems

5.1If the weights are to move at constant velocity, the net force on them must be zero. Thus, the force exerted on the weights is upward, parallel to the displacement, with magnitude 350 N. The work done by this force is

5.2To lift the bucket at constant speed, the woman exerts an upward force whose magnitude is . The work done is , so the displacement is

5.5(a) The force of gravity is given by and is directed downwards. The angle between the force of gravity and the direction of motion is = 90.0° - 30.0° = 60.0°, and so the work done by gravity is given as

(b)The normal force exerted on the block by the incline is , so the friction force is

This Force of friction is directed opposite to the displacement (that is= 180°), and the work it does is

Therefore, Wnet = Wg + Wf = 61.3 J + -46.3 J = 15.0 J

(c)Since the normal force is perpendicular to the displacement; , and the work done by the normal force is .

5.8(a)

(b)

(c)

(d)

5.9(a)The work-energy theorem, , gives
, or

(b), so

5.10Requiring that with , we have
, giving

5.11The person’s mass is . The net upward force acting on the body is . The final upward velocity can then be calculated from the work-energy theorem as

or
which gives

5.15(a)The final kinetic energy of the bullet is

(b)We know that , and also .
Thus,