An electric motor is used to pull a 125 kg box across a floor using a long cable. the tension in the cable is 350N and the box accelerates at 1.2m/s^2[forward] for 5.0s. The cable breaks and the box slows down and stops.

Question

b) How far does the box travel up to the moment the cable breaks?
c)How far does the box travel from the moment the cable breaks until it stops?

Answer 1

a) (1/2) a t^2, where
a = 1.2 m/s^2
t = 5.0 s

b) Speed at cable breakage = a t
V' = 6 m/s

Let distance travelled after cable breakage be X.

M g X *Uk = (M/2) V'^2

You need the coefficient of kinetic friction Uk. You can get it from the initial cable tension, before it breaks.
350 - M*g*Uk = M*a
Uk = (350 - M a) /(Mg)
= 0.163

Now solve for X.

Answer 2

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