Question

A box of cereal is pushed with an applied force of 2.4 N (W) across a table. The cereal box has a mass of 2.5 kg and the box accelerates at 0.41 m/s^2 (W). What is the force of friction? Show your work. What is the coefficient of friction for the table top? Show your work.

Answer 1

What I don't understand is how can the acceleration of the object be 0.41 ms/s when the applied force is only 2.4N and the mass is 2.5kg. In order to get to an acceleration of 0.41ms/s the Force applied must be at least 10.055N. I might be totally forgetting something here or the wording of the equation has thrown me. Can you just check to see if you've written the right figures down please? 2.4N seems pretty low? Cheers
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Answer 2

@Dan.Harrison I'm positive the numbers are correct, but I do agree with you that something seems funny. This is how I solved it, not sure if it's right though:
Fnet= ma
Fnet= 2.5*0.41
Fnet= 1.025

Fnet= Fapp+Ffriction
Ffriction= Fnet-Fapp
Ffriction= 1.025-2.4
Ffriction= -1.375

So the Force of friction is 1.375 N [E]
Yes? No?

Answer 3

I've just realised ive been multiplying mass by 9.81 instead of just leaving it as mass, knew I was missing forgetting something! Your method seems correct, you can calculate the friction coefficient now using f=mu R. Thanks for helping me out!

Answer 4

One last thing, I calculated the coefficient of friction to be:
F=mg
F= 2.5*9.81
F= 24.525 N
-1.375= U * -24.535
U= 0.056
So the coefficient of friction is 0.056?
Thanks a bunch!

Answer 5

I just got 0.056 too, seems reasonable.