What is the mass of the block of ice?

Question

A dockworker applies a constant horizontal force of 77.0N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 11.0m in a time 5.30s .

Answer 1

F=ma
d=1/2 a t^2

therefore

a=2d/t^2

m=F/a

m=F(t^2)/2d

m=77*(5.3)^2/(2*11)

=98 kilos

Answer 2

This is a force-motion problem. F = ma; where m is the mass you are looking for and F = 77 N. (kg-m/sec^2) So we need to know a, the acceleration, to find m = F/a.

For that, we invoke S = 1/2 at^2; where S = 11 m and t = 5.3 sec. In which case, a = 2S/t^2.

Then we plug in m = F/a = F//2S/t^2 and, as everything on the RHS of the = is given, you can solve for m.