A crate of mass 56 kg is pushed across a level floor by a person. If the person exerts a force of 24 N in the horizontal direction and moves the crate a distance of 13 m, what is the work done by the person?

Question

anonymous · Answer

If the crate is starting from rest, use
$$a=\frac{ F }{ m }$$
to find the acc. of the crate, then use $$v^{2}-u^{2}=2as$$
which simplifies to $$v^{2}=2as$$
Thus calculate velocity and then kinetic energy
$$KE=\frac{ mv^{2} }{ 2 }$$
Now according to Work-Energy Theorem, 
Work Done = Change in Kinetic Energy
W=KE(final)-KE(intial)
KE(final), you just calculated using the above formula
KE(initial) will be zero as box was at rest before(v=0 at rest, thus KE=0 at rest)
$$W=\frac{ mv^{2} }{ 2 }-0$$
$$W=\frac{ mv^{2} }{ 2 }$$