Question

A 60 kilogram student jumps down from a laboratory counter. At the instant he lands on the floor hus speed is 3 meters per second. If the student stops in .2 second what is thw average force of floor on the student

Answer 1

As per Newton's law rate of change in momentum is net force

so we can write it as

$F = \frac{dP}{dt}$

$F = \frac{m(v_f - v_i)}{\Delta t}$

now we know that

$m = 60 kg$

$v_f = 3 m/s$

$v_i = 0$

$\Delta t= 0.2 s$

from above equation

$F = \frac{60(3 - 0)}{0.2} = 900 N$

so he will experience 900 N force in above case