Question

A baseball player sliding into second base has an initial velocity of 8.0m/s. The mass of the player is 60.kg. If he comes to rest in a distance of 8.0m, what is the force needed to stop the player?

Answer 1

The equation to obtain the final speed of the player is,

$v^2=u^2+2as$

Substitute the known values,

$\begin{gathered} (0m/s)^2=(8.0m/s)^2+2a(8.0\text{ m)} \\ a=-\frac{64.0m^2s^{-2}}{16.0\text{ m}} \\ =-4m/s^2 \end{gathered}$

The force which is required to stop the player is,

$F=ma$

Plug in the magnitude of known values,

$\begin{gathered} F=(60kg)(4.0m/s^2)(\frac{1\text{ N}}{1kgm/s^2}) \\ =240\text{ N} \end{gathered}$

Thus, the force required to stop the player is 240 N.