Question

You have a level track. You push a cart with mass = 0.84[kg]. You measure the initial velocity to be 0.7[m s-1]. 2 seconds later, you measure the velocity to be 0.525[m s-1]. What is the work (reported in mJ) that friction did on the cart?

work = [mJ]
Use three sig. figs. or N/A if not enough information is given, such as change in position!
Remember, that if the system gains energy, then positive work is done;
otherwise the system loses energy ? work done on system < 0.

Answer 1

1/2mass(velocity final^2- velocity initial^2)

=-90mj

Answer 2

work done by friction=0.5m(v_f²-v_i²)

=0.5*0.84(0.525^2-0.7^2)

=-0.09j

Answer 3

a)Work done by friction=Gain in kinetic energy=0.5mv^2-0.5mu^2=0.5*0.84*0.525^2-0.5*0.84*0.7^2=-0.09J=-90.0mJ