# Work/Energy vs. Force Questions

1. Feb 7, 2009

A couple of questions:

1. Why can't we tell what force a 10 N rock exerts on the ground when dropped from 10 m?
If a car is moving at constant velocity, it's net force is 0, but it is still doing work. How?

2. Work = force * distance, Kinetic energy = 1/2 mass * velocity2, Force * distance = 1/2 mass * velocity2

3. For the first question, I think a valid solution would be that we don't know the acceleration because the problem doesn't tell us to assume we're on earth. However, I think that's kind of a cop out. A more realistic answer might be that we don't know the distance over which the rock exerts force on the ground: The work is 100 J, and if divided by the distance, we would get the force, right?
For the second question, is this a question about what system we're talking about? Because if the car is the system, it is doing exerting force, but if our system is the car and the road, then the car's force and the frictional force cancel for a net force of zero, allowing the car to move at a constant velocity. So if the net force is 0 N, then the work (Force * distance) should be 0 J. But clearly the car is doing work, because it's kinetic energy is 1/2 mass * velocity2. I'm not exactly sure how to reconcile the two equations.

Thanks!

Last edited: Feb 7, 2009
2. Feb 7, 2009

### Nabeshin

Welcome to PF!

For question two, I think you're onto something. There is a frictional force which is preventing the car from just "coasting" at a constant velocity. The net work done on the car is certainly zero because there is no change in kinetic energy, but the car and friction (wind, tires, whatever) both individually do work!

3. Feb 8, 2009

### timmay

For your first question, the force that the rock exerts on the earth will be equal to the acceleration (strictly speaking deceleration but I'll use acc. throughout) it experiences multiplied by its mass. In an ideal system, the acceleration will be constant. The distance that the resulting constant decelerating force acts before the rock finishes deforming the ground will be dependent on the initial potential energy, assuming no losses due to air resistance. You can make a basic assumption that object's GPE will be equal to its kinetic energy upon impact, and that kinetic energy must be input to the earth before the object comes to rest.

So, let's say that your 1 kg rock hits the ground from a height of 10 m. Let's assume it decelerates constantly at 100 g - that is, 100 times the acceleration due to earth's gravitational field.

The rock has an initial potential energy W:

$$W = mgh = 1 kg \times 9.81 ms^{-2} \times 10 m = 98.1 J$$

The decelerating force acting on the rock is given by:

$$F = ma = 1 kg \times 100 \times 9.81 ms^{-2} = 981 N$$

Knowing that the kinetic energy of the rock (equal to the potential energy as it falls) must be transferred to the ground before the rock comes to rest:

$$mgh = \int^{x=X}_{x=0} F dx = 981 N \times X$$

Rearranging:

$$X = \frac{98.1 J}{981 N} = 0.1 m$$

So the object would come to rest after 100 mm in this case.

Unfortunately most impacts involve non-constant accelerations, so to predict the force an object will transmit to another object is pretty difficult without making some big assumptions and understanding how the two objects deform.

Hope this helps