Question

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1. Your car has run out of gas. Fortunately, there is a gas station nearby. You must
exert a force of 700 N on the car in order to move it. By the time you reach the
station, you have done 25000 J of work. How far have you pushed the car?
2. If a neighbor pushes a lawnmower 2 times as far you but exerts only half the
force, which one of you does more work and by how much?
3. A shopper in a supermarket pushes a cart with a force of 40 N directed at an
angle of 30° downward from the horizontal. Find the work done by the shopper on
the cart as the shopper moves along a 45.0 m length of an aisle.
4. If 3.0 J of work is done in raising a 200 g orange, how far is it lifted?
5. A 0.08 kg ball in a kinetic sculpture moves at a constant speed along a motorized
vertical conveyor belt. The ball rises 1.5 m above the ground. A constant frictional
force of 0.3 N acts in the direction opposite the conveyor belt's motion. What is the
net work done on the ball?

Answer 1

1..The car was pushed a distance of 35.71 meters.

2..The neighbor who exerts more force does more work, but the difference in work depends on the specific values of force and distance.

3..The work done by the shopper is 1419 J.

4..The orange is lifted a height of 0.015 meters.

5..The net work done on the ball is 1.176 J.

Step-by-step explanation:

1..When the force of 700 N is exerted on the car, work is done on the car equal to the force times the distance. Let's call the distance pushed "d". Therefore, 700 N x d = 25000 J. Solving for "d" gives a distance of 35.71 meters that the car was pushed to reach the gas station.

2..Work is equal to force times distance, so the neighbor who exerts more force will do more work. Let's call the force and distance that you exert "F1" and "d1", respectively, and the force and distance that the neighbor exerts "F2" and "d2", respectively. If F2 = 0.5F1 and d2 = 2d1, then the work done by the neighbor is (0.5F1)(2d1) = F1d1, which is the same as the work you do. Therefore, both of you do the same amount of work.

3..The work done by the shopper is equal to the force times the distance times the cosine of the angle between the force and the displacement. Let's call the angle between the force and the displacement "θ". Therefore, the work done is W = 40 N x 45.0 m x cos(30°) = 1419 J.

4..The work done in raising the orange against gravity is equal to the gravitational potential energy gained by the orange. Therefore, mgh = 3.0 J, where "m" is the mass of the orange, "g" is the acceleration due to gravity, and "h" is the height that the orange is lifted. Solving for "h" gives a height of 0.015 meters.

5..The net work done on the ball is equal to the change in the ball's kinetic energy. Since the ball moves at a constant speed, its kinetic energy remains constant, so the net work done on the ball is zero. However, since there is a frictional force acting on the ball, work is done against the force of friction. Let's call the distance that the ball rises "h". Therefore, the work done against friction is W = Ff x h = (0.3 N)(1.5 m) = 0.45 J. The work done by the conveyor belt on the ball is equal in magnitude but opposite in direction, so the net work done on the ball is 0.45 J - (-1.626 J) = 1.176 J.