How Does Friction Affect Work Done on an Incline?

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SUMMARY

The discussion focuses on calculating the work done on a crate being pulled up a rough incline, specifically addressing the change in kinetic energy. Key equations include the work done by gravity (Wg = -mgdsin(theta)), the work done by the applied force (W = Fd), and the relationship between kinetic energy and work (Vf = sqrt(2 * change K / m + Vo^2)). The user successfully calculated the work done by gravity as -393.91 J and the work done by the applied force as 1021.36 J but requires assistance in determining the change in kinetic energy and the normal force to calculate friction.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with work-energy principles
  • Knowledge of trigonometric functions in physics
  • Basic grasp of friction and normal force concepts
NEXT STEPS
  • Calculate the normal force on an incline using N = mgcos(theta)
  • Determine the frictional force using f = coefficient of friction * N
  • Apply the work-energy principle to find the change in kinetic energy
  • Explore the relationship between work done and kinetic energy in various scenarios
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Students studying physics, particularly those focusing on mechanics and energy concepts, as well as educators looking for practical examples of work and energy calculations on inclined planes.

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

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