## Work Energy Theorem with Kinetic Friction and External Work

1. The problem statement, all variables and given/known data

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.5 m/s. The pulling force is 100 N parallel to the incline which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is .4 and the cart is pulled 5.00 m.

e) What is the speed of the crate after being pulled 5.00m?

2. Relevant equations
Delta Energy Mechanical = -Force_friction(d) + Work_external

K_f + U_f = K_i + U_i - Force_friction(d) + Work_external

For constant Force and parallel force and displacement: W = F(d)

3. The attempt at a solution

K_f + U_f = K_i + U_i - Force_friction(d) + Work_external

.5mvf2 + mgh = .5mvi2 + 0 - Force_friction(d) + Work_external

h = 5sin(20°)
Force_friction = -μ Fn
= -μ 98cos(20°)
Work External = F(d) = 100N * 5m = 500J

.5mvf2 + mgh = .5mvi2 +μ 98cos(20°)(5) + 500

vf2 = (.5mvi2 +μ 98cos(20°)(5) + 500 - mgh) / (.5m)

v = 10.27 m/s

The answer is given as 5.65 m/s.
Any help is appreciated!
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 Quote by kc0ldeah vf2 = (.5mvi2 +μ 98sin(20°)(5) + 500 - mgh) / (.5m)
Is that a typo, or did you change cosine into sine?
 Sorry that was a typo I will fix it.

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## Work Energy Theorem with Kinetic Friction and External Work

 Quote by kc0ldeah .5mvf2 + mgh = .5mvi2 +μ 98cos(20°)(5) + 500
Check the sign of the work done by friction.

 Quote by Doc Al Check the sign of the work done by friction.
K_f + U_f = K_i + U_i - Force_friction(d) + Work_external

If you plug -μ98cos(20°) into that it should be +μ98cos(20°) right?
It is -μ98cos(20°) to begin with because Force_friction acts in the opposite direction of the crate's movement.

EDIT: But fixing it the way you said gives me 5.65 which is the answer. Why is that term supposed to be negative?

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 Quote by kc0ldeah K_f + U_f = K_i + U_i - Force_friction(d) + Work_external If you plug -μ98cos(20°) into that it should be +μ98cos(20°) right? It is -μ98cos(20°) to begin with because Force_friction acts in the opposite direction of the crate's movement. EDIT: But fixing it the way you said gives me 5.65 which is the answer. Why is that term supposed to be negative?
It depends on how you define the terms. That friction term represents the work done by friction, which is negative (since friction opposes the displacement). When you wrote "-Force_friction(d)", I assume you meant for Force_friction to be the magnitude of that force.

More generally: Work done by friction (or any force) = Force*distance = (-μmgcosθ)(d)

Final Energy = Initial Energy + Work done by friction + Work done by applied force
 Final Energy = Initial Energy + Work done by friction + Work done by applied force Thank you this is what I really needed to see. I guess the equation my teacher gave me already factored in the fact that friction is an opposing force. That has really been messing me up.

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 Quote by kc0ldeah I guess the equation my teacher gave me already factored in the fact that friction is an opposing force.
Exactly.