A 40 kg gymnast somersaults into a foam ball pit at a speed of 7 m/s. If the foam applies an average resistive force of 1,000 N,

Question

2 years ago

4

how far into the pit will the gymnast sink before she stops?

1 answer:

enot [183] · Answer 1 · 2023-10-01T13:27:42+03:00

The distance covered is 0.98 m

Explanation:

Newton's second law states that the force applied on the gymnast is equal to the product between its mass and its acceleration:

$F=ma$

where in this case,

F = -1000 N is the force applied (negative since it is opposite to the direction of motion)

m = 40 kg is the mass

a is the acceleration

Solving for a,

$a=\frac{F}{m}=\frac{-1000}{40}=-25 m/s^2$

Since the motion of the gymnast is a uniformly accelerated motion, we can now apply suvat equations:

$v^2-u^2=2as$

where

v = 0 is the final velocity of the gymnast

u = 7 m/s is the initial velocity

$a=-25 m/s^2$ is the acceleration

s is the distance through which the gymnast moves before stopping

And solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{0-7^2}{2(-25)}=0.98 m$

Learn more about accelerated motion:

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