# Practice Test- Box sliding down a slope

**1.**A 30kg box is placed on a 15° slope and a person pushes on the box up the slope to keep it from sliding. If the coefficient of static friction is 0.15, what is the minimum force that the person must exert to keep the box in place?

Answer Choices:

A. 273N

B. 241N

C. 33.5 N

D. 64.7 N

E. 294 N

**2.** Fs≤μs*N

F=μ*N

N=mg <—–not sure about that because I know normal force changes depending on the situation but I think this applies here.

Also my teacher has us use g= 9.8 NOT 9.81

**3. **So first I calculated the normal force: N=mg —–> (30)(9.8) = 294 N

Then I put my normal force into the equation F=μ*N ——> (.15)(294)= 44.1 N

This was not one of the answer choices. However I recalled this equation : Fs≤μs*N.

Since I was solving for the force that was a minimum I assumed that out of my answer choices C. 33.5 N was correct because 33.5 ≤ 44.1 which was my calculated force. This turned out to be the correct answer on the answer sheet. However I feel that this was more of a guess that happened to be right rather than the correct way of doing this.

Any help is appreciated!

http://ift.tt/1pX118R

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