Question

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by 0.40 m, the mass that moves downward is 61.0 kg, and the collision on the floor lasts 0.0820 s. What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

Answer 1

Given the relationships,

$h=0.4m\\m=61kg\\t=0.082s$

Given the energy conservation equation,

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

$v= \sqrt{2gh}$

$v= \sqrt{2*9.8*0.4}$

$v=2.8m/s$

Since this is the initial speed and according to the problem, the final is zero,

\vec{J}=m(\vec{v}_f-\vec{v}_i)

$\vec{J}=61*(0-(-2.8))$

$\vec{J}=170Kg-m/s$

B) The average force would be given by the change that exists at the time during the change of time,

$|F|=\frac{P_f-P_i}{\Delta t}$

$|F|= \frac{m(0-v_i)}{\Delta t}$

$|F| = \frac{170}{0.082}$

$|F|= 2073.1N$