Answer to Question #111497 in Mechanics | Relativity for Jason Grace

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, what is the percent of mechanical energy lost to thermal energy due to friction?

Answer 1

All the energy of the bullet is kinetic energy, passing through the wooden block part of this energy is lost and converted into thermal energy now the total energy consists of a part of kinetic energy and heat, which is scattered over the mass of the block, mathematically it looks like this

$E_k=E'_k+Q\\\frac{m\times{ v^2}}{2}=\frac{m\times{ v^2}'}{2}+Q;\\Q=\frac{m}{2}\times(v^2-{v^2}');\\Q=\frac{3.25\times10^{-3}kg}{2}\times(80^2\frac{m^2}{s^2}-{20^2\frac{m^2}{s^2}})=9.75J;$

Energy of the bullet before entering the block:

$\frac{m\times{ v^2}}{2}=\frac{3.25\times10^{-3}\times{6400}}{2}=10.4J;$

The percent is

$\frac{9.75\times100}{10.4}=93.75\%$

Answer: 93.75% of kinetic energy is converted to heat