Question

A 3.250 g bullet enters a wooden block with a velocity of 80.00 m/s. If it exits the block with a velocity of 20.00 m/s, find the percent of mechanical energy lost to thermal energy due to friction.

Answer 1

loss in mechanical energy = loss in kinetic energy =$\frac12m(v_1^2-v_2^2)=\frac12\times3.25\times10^{-3}(80^2-20^2)=9.75 J$

initial kinetic energy = $\frac12mv_1^2=10.4J$

loss percent = $\dfrac{9.75}{10.4}\times 100=93.75\%$