Question

An 80 kg man drops to a concrete patio from a window 0.50m above the patio. He neglects to bend his knees on landing, taking 2.0cm to stop.
(a) What is his average acceleration from when his feet first touch the patio to when he stops?
(b) What is the magnitude of the average stopping force exerted on him by the patio?

Answer 1

Hint: Here the mass of the man is given and he drops to a concrete patio, so, his initial velocity is zero. The height is 0.5m and after striking the patio it takes 2 cm more to stop. So, the patio exerts a force on the man because man also exerts a force on the patio following Newton’s third law.

Complete step by step answer:
Mass, m= 80 kg
Initial velocity, u= 0
Height, h= 0.5 m
Using third equation of motion to find the velocity of the man striking the patio,
$$\begin{array}{rl}& \Rightarrow v^2-u^2=2as\\ & \Rightarrow v^2-0=2\times 9.8\times 0.5\\ & \Rightarrow v^2=9.8\\ & \therefore v=3.13m/s\end{array}$$
So, when the man strikes the ground his velocity is 3.13 m/s. Now he moves 2 cm more that is 0.02 m and finally comes to rest. Using
$$\begin{array}{rl}& \Rightarrow v^2-u^2=2as\\ & \Rightarrow 0-{3.13}^2=2\times a\times 0.02\\ & \therefore a=-\frac{9.8}{0.04}=-245m/s^2\end{array}$$
Acceleration is negative because the man comes to rest.
So, the average acceleration from when his feet first touch the patio to when he stops is 245 $$m/s^2$$.
(b) Now in order to find force using Newton’s second law
$$\begin{array}{rl}& \Rightarrow F=ma+mg\\ & \Rightarrow F=m(g+a)\\ & \Rightarrow F=80(9.8+245)\\ & \Rightarrow F=80\times 254.8\\ & \therefore F=20384N\end{array}$$

So, the value of average force exerted is 20384 N.

Note: Here in part (B) while finding out the magnitude of average force being exerted, we have taken both the forces into account one is due to motion and the other is due to gravity. Newton’s law should be applied carefully by taking into account the directions.