A dock worker applies a constant horizontal force of 79.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in the first 6.00 s.
What is the mass of the block of ice?
Work done = final kinetic energy
F X = (1/2) M V^2
The final velocity V is twice the average velocity of 11/6 m/s, so
V = 22/6 = 3.67 m/s
Solve for M
M = 2 F X/(V^2)
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11.7
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