A dockworker applies a constant horizontal force of 80. 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 5. 00 s. (a) what is the mass of the block office? (b) If the worker stops pushing at the end of 5. 00 s, how far does the block move in the next 5. 00 s?.

Explanation:

Let's start by finding the acceleration of the block. We know that its initial velocity is 0 and also the initial displacement. Therefore we can write:

s = 1/2·a·t²

We know that s = 11 m and t = 5s. The acceleration will be:

a = (2s)/t² = 22m/25s² = 0,88 m/s²

From the relation F=m·a we can find the mass:

m = F/a = 80N/0.88m/s² ≈ 90.9 kg.

After the worker stops pushing the block, its velocity will be:

v = a·t = (0,88 m/s²)·5s = 4,4 m/s

Therefore in the next 5 seconds it will travel a distance:

s = v·t = 4,4 m/s · 5s = 22m

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