Question

A force of 5 N is applied on a 20 Kg mass at rest. The work done in the first second is

Answer 1

acceleration of the block=5/20=0.25
using second equation of motion to find displacement
$s = 0 \times 1 + 0.5 \times 0.25 \times {1}^{2} \\ s = 1 + 0.125 = 1.125m$
so displacement=1.125m
work =force*displacement
work=5*1.125=5.625j
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