Question

21) A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 10.0 ms, what is her momentum just as she hits the floor? What average force acts on her during the deceleration?

Answer 1

Answer:

-22,150 N

Explanation:

When the youngster jumps off the platform, during the fall her initial potential energy is converted into kinetic energy, according to the law of conservation of energy. Therefore, we can write:

$mgh=\frac{1}{2}mu^2$

where the term on the left is the potential energy while the term on the right is the kinetic energy, and where

m = 50.0 kg is the mass of the youngster

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 1.00 m is the heigth of the platform

u is the speed of the youngster as she reaches the floor

Solving for u,

$u=\sqrt{2gh}=\sqrt{2(9.8)(1.00)}=4.43 m/s$

Then, when the youngster hits the floor, the force exerted on her during the deceleration is given by:

$F=\frac{\Delta p}{\Delta t}=\frac{m(v-u)}{\Delta t}$

where $\Delta p$ is her change in momentum, and where

m is the mass

v = 0 is the final velocity (she comes to a stop)

u = 4.43 m/s is the initial velocity

$\Delta t=10.0 ms =0.010 s$ is the duration of the collision

Substituting,

$F=\frac{(50.0)(0-4.43)}{0.010}=-22150 N$

And the negative sign means the direction of the force is opposite to the motion (so, upward).