Question

A dock worker applies a constant horizontal force of 75.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 13.0 m in the first 4.00 s. What is the mass of the block of ice? m= (value) Kg

Answer 1

Using Net force balance on ice block

F_net = m*a

m = F_net/a

F_net = 75.0 N

a = acceleration of block = ?

Using 2nd kinematic equation:

d = U*t + (1/2)*a*t^2

d = 13.0 m, U = 0 m/sec and t = 4.00 sec, So

13.0 = 0*4.00 + (1/2)*a*4.00^2

a = 13.0*2/4.00^2 = 1.625 m/sec^2

So,

m = F_net/a = 75.0/1.625

m = 46.15 kg = mass of block of ice

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