How Does Friction Affect Work Done on an Incline?

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The discussion revolves around calculating the work done on a crate being pulled up a rough incline, focusing on gravitational force, applied force, and friction. The work done by gravity is calculated as -393.91 J, indicating it is negative. The work done by the pulling force is noted to be 1021.36 J, but there is confusion regarding whether the pulling force is 126 N or 136 N. The participant seeks assistance in determining the change in kinetic energy, which involves understanding the work done against friction and the normal force. The overall goal is to find the speed of the crate after being pulled 7.51 m, emphasizing the importance of accurately calculating the forces involved.
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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?
 
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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?
 
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