Work Done on a Suitcase by Various Forces

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SUMMARY

The discussion focuses on calculating the work done on a suitcase by various forces while being pulled up a ramp. The suitcase has a mass of 16.7 kg, and the applied force is 155 N at a 30-degree incline. The coefficient of kinetic friction is 0.267, and the suitcase travels 4.10 m. The total work done on the suitcase is calculated as 371 J, factoring in the work done by the applied force and friction, while the normal and gravitational forces contribute zero work due to their perpendicular orientation to the displacement.

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Homework Statement



*I already figured the problem out*

A luggage handler pulls a suitcase of mass 16.7 kg up a ramp inclined at an angle 30.0 degrees above the horizontal by a force \overline{F} of magnitude 155 N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is 0.267. The suitcase travels a distance 4.10 m along the ramp.

A. Calculate the work done on the suitcase by the force F.
B. Calculate the work done on the suitcase by the gravitational force.
C. Calculate the work done on the suitcase by the normal force.
D. Calculate the work done on the suitcase by the friction force.
E. Calculate the total work done on the suitcase.

F. If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled 4.10 m along the ramp?

Homework Equations



F*s = Work

The Attempt at a Solution



A. Is it:
work of F applied= (155 N)(4.10 m)cos30.0 = 550.0 J

B. Is it
work of friction=(43.7 N)(4.10 m)cos180.0 = -179.0 J

C and D. Wouldn't both be 0, considering cos90 = 0?

E. Is it just 550 J + 0 J + 0 J + -179 J = 371 J?

F. No idea, need help.

Thoughts? Thanks!
 
Last edited:
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s is along the ramp right? So, why are you projecting things to the horizontal direction?
 
So work of f applied= 155 N(4.10 m)?

*I figured this problem out*
 
Last edited:

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