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Alabran
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While tutoring a friend of mine, I encountered a problem I was startled to discover I was unable to find the correct answer to. I am far from being a stranger to Work, yet the solution eludes me. I don't remember the exact values of the coefficient of friction et all, so I'll provide the problem in terms of the variables.
A block of mass M is being pulled across a rough surface by a rope at angle theta above the block, providing a tension force T. The coefficient of kinetic friction between the block and the surface is MuK. What is the work done by friction over the distance D.
F(friction) = Mu(k) * N
T(y) = TSin(theta)
W = F * D
Under the understanding that the work is equal to the force multiplied by the distance, we went to solve for the force, first. Force of friction is Mu(k) * N, seemingly simple. So, since there is actively a force pulling up on the block, the block would not be pressing down on the surface as much thus the normal force would be less. We solved for the normal force under the assumption that the N would be equal to the difference between the force of gravity downward, and the upward pull due to tension. N = Fg - Ft. Then, solving for both values, expanded the problem to: N = Mg - TSin(theta). We then plugged this into the friction force equation, F = Mu(k) * N, resulting in F = Mu(k) * (Mg - TSin(theta)). Finally, after finding that, we plugged that force into the equation for the Work: W = F * D. Thus, our final equation was: Mu(k) * (Mg - TSin(theta)) * D. For some reason, this was incorrect, and I am at a lack to explain why that may be. Any help you all could provide would be greatly appreciated.
Thanks
Homework Statement
A block of mass M is being pulled across a rough surface by a rope at angle theta above the block, providing a tension force T. The coefficient of kinetic friction between the block and the surface is MuK. What is the work done by friction over the distance D.
Homework Equations
F(friction) = Mu(k) * N
T(y) = TSin(theta)
W = F * D
The Attempt at a Solution
Under the understanding that the work is equal to the force multiplied by the distance, we went to solve for the force, first. Force of friction is Mu(k) * N, seemingly simple. So, since there is actively a force pulling up on the block, the block would not be pressing down on the surface as much thus the normal force would be less. We solved for the normal force under the assumption that the N would be equal to the difference between the force of gravity downward, and the upward pull due to tension. N = Fg - Ft. Then, solving for both values, expanded the problem to: N = Mg - TSin(theta). We then plugged this into the friction force equation, F = Mu(k) * N, resulting in F = Mu(k) * (Mg - TSin(theta)). Finally, after finding that, we plugged that force into the equation for the Work: W = F * D. Thus, our final equation was: Mu(k) * (Mg - TSin(theta)) * D. For some reason, this was incorrect, and I am at a lack to explain why that may be. Any help you all could provide would be greatly appreciated.
Thanks