Work of Friction Done Opposing Force at an Angle

[awilhite]

Homework Statement

Ben does 600J of work pushing a 55kg cart 5m in the +x direction by applying a force at a downward angle of 30° to the direction of displacement. If the coefficient of kinetic friction μκ is equal to 0.1, find:

• The magnitude of the applied force.
• The work done by the force of friction on the cart.

Homework Equations

W = Fd
Work of friction = -μκ(mg + Fsinθ)d
Work in x direction = Fcosθ

The Attempt at a Solution

The magnitude of the applied force:

W = Fd
600 = F(5)
F = 120N

The work done by friction:

Wopp = -μκ(mg + Fsinθ)d
Wopp = -0.1(55*10 + 120sin30°)(5)
Wopp = -305J



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Is it right? I have an undying feeling that it is correct; however, the student teacher that teaches my class substituted the initial W = Fd with W = Fdcosθ. That would only give you the force in the x direction, and then the force of friction found in the second part would be incorrect, right?

Help me out, please. Thanks.

 

[PhanthomJay]

Teacher is correct, there is no work being done in the y direction because there is no displacemnt in the y direction. For constant force, Work is equal to the product of the force times the displacement in the direction of the force (W=Fdcostheta). (Or you could say, Work is equal to the component of the force in the direction of the displacement time the displacement...W = Fcostheta(d)).