# Work/Energy vs. Force Questions

In summary, the first question discusses why it is difficult to determine the force exerted by a 10 N rock on the ground when dropped from 10 m. This is because the acceleration and distance over which the force is exerted are unknown. The second question delves into the concept of work and how it relates to a car moving at a constant velocity. While the net force may be zero, both the car and frictional forces still do work.
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:
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 = 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!

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!

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

## 1. What is the difference between work and energy?

Work and energy are both physical quantities used to describe the motion of objects. Work is defined as the product of the force applied on an object and the displacement of the object in the direction of the force. Energy, on the other hand, is the ability of an object to do work. It is measured in the same units as work (Joules) and is conserved in a closed system.

## 2. How are work and energy related?

Work and energy are closely related concepts. Work is the transfer of energy from one object to another, or from one form to another. This means that when work is done on an object, its energy changes. Conversely, when an object does work, it is using its energy to achieve a certain result.

## 3. What is the formula for calculating work?

The formula for calculating work is work = force x displacement x cos(theta). In simpler terms, work is equal to the force applied on an object multiplied by the distance the object moves in the direction of the force. The angle between the force and the displacement is represented by theta.

## 4. How does force affect work and energy?

Force is a crucial factor in the equations for work and energy. The magnitude and direction of the force applied on an object determine the amount of work done on the object and the change in its energy. A larger force will result in more work done and a greater change in energy, while a force in the opposite direction of motion will result in negative work and a decrease in energy.

## 5. Can work done on an object be negative?

Yes, work can be negative if the force applied on an object is in the opposite direction of its displacement. This means that the object is losing energy instead of gaining it. For example, if you push a book off a table, the work done by gravity on the book is negative because the force of gravity is acting against the book's displacement.

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