# What is the Work Done by Force F on a Moving Object?

• WahooMan
In summary: You are supposed to consider the vector sum of all the forces, which you have not done. The x component of that sum is 4.0 kg * 5.0 m/s^2 = 20.0 N, and the y component is 4.0 kg*12.0 m/s^2 = 48.0 N. So the net force is (20.0i + 48.0j) N. Now you can find the distance traveled by using the kinematics equations of motion. The work done is (20.0i + 48.0j) N * (distance) = 1518 J. In summary, the problem involves a 4.
WahooMan

## Homework Statement

A 4.0 kg object moving in two dimensions initially has a velocity V1 = (11.0i + 20.0j)m/s. A net force F then acts on the object for 2.0s, after which the object's velocity is V2 = (16.0i + 32.0j)m/s. Determine the work done by F on the object. Enter your answers numerically separated by a comma.

## Homework Equations

W = (1/2)(m)(V2^2 - V1^2)
or
W = (Integral from A to B) of F * dl

## The Attempt at a Solution

I tried the first equation and got

W = (1/2)(4.0)(1280 - 521) = 1518J

but my instructions are to enter my answers (plural) separated by a comma, which I can't figure out because I don't understand why I would have two answers.

Also, I don't quite understand how to do the integral equation. Would A = 0s and B = 2.0s? How do you integrate 13.0 N (what I got for the net force) with respect to l?

Any help at all would be appreciated..

WahooMan said:

## Homework Statement

A 4.0 kg object moving in two dimensions initially has a velocity V1 = (11.0i + 20.0j)m/s. A net force F then acts on the object for 2.0s, after which the object's velocity is V2 = (16.0i + 32.0j)m/s. Determine the work done by F on the object. Enter your answers numerically separated by a comma.

## Homework Equations

W = (1/2)(m)(V2^2 - V1^2)
or
W = (Integral from A to B) of F * dl

## The Attempt at a Solution

I tried the first equation and got

W = (1/2)(4.0)(1280 - 521) = 1518J
but my instructions are to enter my answers (plural) separated by a comma, which I can't figure out because I don't understand why I would have two answers.
Yes, good, this is correct. I don't know what that comma is all about, either.
Also, I don't quite understand how to do the integral equation. Would A = 0s and B = 2.0s? How do you integrate 13.0 N (what I got for the net force) with respect to l?
This is the harder way to do it, but if you are integrating an Fdx term, the integral must be from x =0 to x = x, but since the force is constant, the work done is just F_net(d), where d is the distance traveled in 2 seconds. That involves the use of F_net =ma to find the acceleration, and then using the kinematic equations of motion to find d. Your calc for the net force, however, is incorrect.

## 1. What is work done by a force?

Work done by a force is the product of the force applied on an object and the distance that object moves in the direction of that force. It is a measure of the energy transferred to an object by the force.

## 2. How do you calculate work done by a force?

To calculate work done by a force, you need to multiply the magnitude of the force by the distance the object moves in the direction of the force. This can be represented by the equation W = F * d, where W is the work done, F is the force, and d is the distance.

## 3. What are the units of work done by a force?

The SI unit of work is joule (J), which is equal to 1 newton (N) multiplied by 1 meter (m). Other common units of work include foot-pound (ft-lb) and calorie (cal).

## 4. Can the work done by a force be negative?

Yes, the work done by a force can be negative. This happens when the force and the direction of movement are in opposite directions. In this case, the work done is considered to be negative as the force is acting in the opposite direction of the movement of the object.

## 5. How does the angle between the force and the direction of movement affect the work done?

The angle between the force and the direction of movement affects the work done by the force. When the force and the direction of movement are perpendicular, the work done is equal to zero. As the angle decreases, the work done increases, reaching its maximum when the force is acting in the same direction as the movement.

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