A box slides to the right across a horizontal floor. A person called Ted exerts a force T to the right on the box. A person called Mario exerts a force M to the left, which is half as large as the force T. Given that there is friction f and the box accelerates to the right, rank the sizes of these three forces exerted on the box. a. f < M < T
b. M < f < T
c. M < T < f
d. f = M < T
e. It cannot be determined.

a. f < M < T

Explanation:

Let us take the right direction as positive.

Since Ted exerts a force T to the right his force is +T, Mario exerts a force M to the left, his force is -M. It is also given that Mario's force is half of Ted's force, so M = T/2. Finally, the frictional force , f is to the left, so it is -f. Let the net force be F and it is to the right since the box moves to the right.

So, +T - M - f = +F

Substituting M = T/2, we have

+ T -T/2 - f = F

+ T/2 - f = F

+T/2 = F + f

So, T = 2(F + f) and

M = T/2 = F + f

Since T = 2(F +f) = 2M, It follows that T > M

Also, since M = F + f, it follows that M > f

So, T > M > f ⇒ f < M < T

So, the answer is a.

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