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Question: A 4.04kg box sits at rest at the bottom of a ramp that is 8.62m long and that is inclined at 40.0∘ above the horizontal. The coefficient of kinetic friction is 0.40, and the coefficient of static friction is 0.50.

What constant force F , applied parallel to the surface of the ramp, is required to push the box to the top of the ramp in a time of 3.67s ?

Attempt: I have $a_x=\frac{2(x-x_0)}{t^2}$ and $F=m(a_x+gcos(\theta)+\mu_kmgsin(\theta))$.

So I can combine both equations to have $F=m(\frac{2(x-x_0)}{t^2} + gcos(\theta) + \mu_kmgsin(\theta))$

When I plug in all my values into the the equation above I get 76.7N, however mastering physics is telling me that my answer is wrong. Am I or not?

Correct: $a_x=\frac{2(x-x_0)}{t^2}$ and $F=m(a_x+gsin(\theta)+\mu_kgcos(\theta))$.

So I can combine both equations to have $F=m(\frac{2(x-x_0)}{t^2} + gsin(\theta) + \mu_kgcos(\theta))$

Thus the answer is 43N.

Username Unknown
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1 Answers1

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You appear to have your trig functions backwards. The angle given is the incline above the horizontal; if the angle is zero, then gravity should be perpendicular to the surface, and thus should not contribute directly to the necessary force (it would contribute only through friction). But $\cos(0)=1$, so you have it at maximum in this case.

Try switching $\cos$ and $\sin$ in your force equation. Does this produce the given answer?

There's another issue: your friction force, as written, is $\mu_k m^2 g\sin(\theta)$. Aside from the trig function being wrong, there's an $m^2$, which doesn't make sense.

Glen O
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