## College Physics (4th Edition)

The kinetic energy of Charles Murphy and his bicycle is $27,200~J$ The kinetic energy of John Howard and his bicycle is $1.63\times 10^5~J$
We can find the kinetic energy of Charles Murphy and his bicycle: $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}(70.5~kg)(27.8~m/s)^2$ $KE = 27,200~J$ The kinetic energy of Charles Murphy and his bicycle is $27,200~J$ We can find the kinetic energy of John Howard and his bicycle: $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}(70.5~kg)(68.04~m/s)^2$ $KE = 1.63\times 10^5~J$ The kinetic energy of John Howard and his bicycle is $1.63\times 10^5~J$.