How Does Friction Affect Work Done on an Incline?

[BitterSuites]

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?

 

[Doc Al]

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?

 

[Moetasim]

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!