Question

a box of mass 5.0 kris accelerated from rest across a floor at a rate of 2.0 m/s^2 for 7.0 s. what was the work done on the box.

Answer 1

\[W=Fd\]where W=work, F=force on the box and d=displacement. Now, F=ma thus F=5*2=10. Now to get displacement we use s=1/2at^2 so, s=(1/2)(2)(7)^2=49 meters. So the work done is W=10*49=490 joules.

Answer 2

If the box is moving on a frictionless surface, then use work energy theorem; work done = change in K.E.
Initial Kinetic Energy=0 as the object is at rest.
using v=u+at , v=2x7=14m/s
Then Change in K.E=1/2(5)[14^{2}\]= 490J0ule