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Work Along An Incline

by BitterSuites
Tags: incline, work
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BitterSuites
#1
Mar4-08, 01:10 PM
P: 38
1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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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Doc Al
#2
Mar4-08, 03:16 PM
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P: 41,440
Quote Quote by BitterSuites View Post
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.

2. Relevant equations

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

3. 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 = [itex]\mu N[/itex]. What's the normal force here?


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