# Work Along An Incline

• BitterSuites
In summary, a crate is pulled up a rough incline with a pulling force parallel to the incline for a distance of 7.51m. The acceleration of gravity is 9.8 m/s^2 and the given values are d=7.51, theta=32, m=10.1, g=9.8, coefficient of friction=.178, and v=1.77. We have solved for the magnitude of work done by the gravitational force, which is -393.91 J, and determined that it is negative. The work done by the 126 N force is 1021.36 J. However, we are stumped on finding the change in kinetic energy of the crate and need help

## Homework Statement

A crate is pulled up a rough incline. The pulling force is parallel to the incline. The crate is pulled a distance of 7.51m.

The acceleration of gravity is 9.8 m/s^2.

Given:

d = 7.51
theta = 32
m = 10.1
g = 9.8
coefficient of friction = .178
v = 1.77

1. What is the magnitude of the work done by the gravitational force?
2. Is the work done by the gravitational field zero, positive, or negative?
3. How much work is done by the 126 N force?
4. What is the change in kinetic energy of the crate?
5. What is the speed of the crate after it is pulled 7.51 m?

I have solved for 1-3 and know how to do 5, but need 4 in order to do it. I will show all work below, but specifically need help with #4.

## Homework Equations

wg = -mgdsintheta
w=Fd
Vf = sqrt 2 * change K / m + Vo^2

## The Attempt at a Solution

1. Wg = -mgdsintheta = -10.1*9.8*7.51sin32 = -393.91 J

|Wg| = 393.91

2. As shown in #1, it is negative.

3. W = Fd = 136 * 7.51 = 1021.36

4. I'm stumped. I think Wf = -fd but I can't remember how to get f. My brain shut off half way into #3. I think Wapplied matters as well, but I'm drawing a blank.

Change in K would equal Wg + Wapplied + Wf (I think)

5. Vf = sqrt 2 * change K / m + Vo^2

Anyone mind taking my hand and walking me through this one?

BitterSuites said:
Given:

d = 7.51
theta = 32
m = 10.1
g = 9.8
coefficient of friction = .178
v = 1.77
What's v?

1. What is the magnitude of the work done by the gravitational force?
2. Is the work done by the gravitational field zero, positive, or negative?
3. How much work is done by the 126 N force?
Is this the "pulling force"?
4. What is the change in kinetic energy of the crate?
5. What is the speed of the crate after it is pulled 7.51 m?

I have solved for 1-3 and know how to do 5, but need 4 in order to do it. I will show all work below, but specifically need help with #4.

## Homework Equations

wg = -mgdsintheta
w=Fd
Vf = sqrt 2 * change K / m + Vo^2

## The Attempt at a Solution

1. Wg = -mgdsintheta = -10.1*9.8*7.51sin32 = -393.91 J

|Wg| = 393.91

2. As shown in #1, it is negative.

3. W = Fd = 136 * 7.51 = 1021.36
Is the force 126 or 136N?
4. I'm stumped. I think Wf = -fd but I can't remember how to get f. My brain shut off half way into #3. I think Wapplied matters as well, but I'm drawing a blank.

Change in K would equal Wg + Wapplied + Wf (I think)
Good. Kinetic friction = $\mu N$. What's the normal force here?

Sure, I'd be happy to help you with #4. To find the change in kinetic energy, we need to know the initial kinetic energy and the final kinetic energy. The initial kinetic energy can be calculated using the formula KE = 1/2 * m * v^2, where m is the mass of the crate and v is the initial velocity (which is 0 in this case). So, KEi = 1/2 * 10.1 * 0^2 = 0 J.

Next, we need to calculate the final kinetic energy. We know that the crate is pulled a distance of 7.51 m, so we can use the formula W = Fd to find the work done by the applied force. We found this in #3 to be 1021.36 J. Now, we also need to take into account the work done by the gravitational force, which we found in #1 to be -393.91 J. The negative sign indicates that the work done by gravity is in the opposite direction of the movement, so it is actually taking away kinetic energy from the crate. Lastly, we need to consider the work done by friction, which is equal to the coefficient of friction times the normal force times the distance moved. The normal force is equal to the weight of the crate, which we found to be 10.1 * 9.8 * cos32 = 85.71 N. So, the work done by friction is -0.178 * 85.71 * 7.51 = -101.54 J. Again, the negative sign indicates that this work is also taking away kinetic energy from the crate.

Now, to find the final kinetic energy, we can add all of these values together: KEf = 0 + 1021.36 - 393.91 - 101.54 = 526.91 J.

So, the change in kinetic energy is given by KEf - KEi = 526.91 - 0 = 526.91 J. This means that the kinetic energy of the crate has increased by 526.91 J as a result of being pulled up the incline. I hope this helps guide you through the problem!

## What is work along an incline?

Work along an incline refers to the physical effort required to move an object up or down an inclined plane. It involves exerting a force in a direction parallel to the incline, rather than perpendicular to it.

## How is work along an incline different from work on a flat surface?

The main difference is the direction in which the force is applied. On a flat surface, the force is perpendicular to the surface, while on an incline, it is parallel to the surface. This requires more effort as the force must overcome the component of gravity acting against it.

## What factors affect the amount of work along an incline?

The amount of work along an incline is affected by the angle of the incline, the weight of the object being moved, and the force applied. The steeper the incline and the heavier the object, the more work is required to move it along the incline.

## How can work along an incline be calculated?

Work along an incline can be calculated using the formula W = Fdcosθ, where W is work, F is force, d is displacement, and θ is the angle of incline. This formula takes into account the component of the force parallel to the incline.

## What are some real-life examples of work along an incline?

Some examples include pushing a shopping cart up a ramp, pulling a sled up a hill, or rolling a heavy object up a loading ramp. Inclines are commonly used in transportation, construction, and sports to make work easier and more efficient.

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