Work-Energy Theorem EX 1

(a) Using the relationship for the change of kinetic energy $\Delta K=K_{f}-K_{i}={\frac {1}{2}}m{v_{f}}^{2}-{\frac {1}{2}}m{v_{i}}^{2}$, we find

$\Delta K={\frac {1}{2}}(0.1)(18)^{2}-{\frac {1}{2}}(0.1)(20)^{2}=-3.8J$

(b) The work (or more precisely the net work) done is equal to the change in kinetic energy:

$W=\Delta K=-3.8J$