Work-Energy Theorem EX 1

(a) Using the relationship for the change of kinetic energy $\mathrm{\Delta }K=K_f-K_i=\frac{1}{2}m{v_f}^2-\frac{1}{2}m{v_i}^2$, we find

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

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

$W=\mathrm{\Delta }K=-3.8J$