Answer
The truck's kinetic energy increases by $~~74,600~J$
Work Step by Step
We can convert $41~km/h$ to units of $m/s$:
$(41~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 11.39~m/s$
We can convert $51~km/h$ to units of $m/s$:
$(51~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 14.17~m/s$
We can find the initial kinetic energy:
$K_i = \frac{1}{2}mv^2$
$K_i = \frac{1}{2}(2100~kg)(11.39~m/s)^2$
$K_i = 136,200~J$
We can find the final kinetic energy:
$K_f = \frac{1}{2}mv^2$
$K_f = \frac{1}{2}(2100~kg)(14.17~m/s)^2$
$K_f = 210,800~J$
We can find the change in kinetic energy:
$\Delta K = K_f-K_i$
$\Delta K = 210,800~J-136,200~J$
$\Delta K = 74,600~J$
The truck's kinetic energy increases by $~~74,600~J$.
