A 0.5 kg block initially at rest on a frictionless, horizontal surface is acted upon by a force of 5.0 N for a distance of 4.0 m. How much kinetic energy does the block gain? What is its final velocity?

Answer 1

mass=0.5 kg
force=5 N
f=ma
a=force/mass
a=5/0.5
a=10m/s^2
u=0
s=4m
v^2-u^2=2as
v^2=2*10*4
v^2=80
final velocity = v = 8.9m/s
kinetic energy = 0.5*m*v^2=0.5*0.5*80
=20
gain in kinetic energy= final kinetic energy -initial kinetic energy
=20-0=20