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 63.0 kg, and the collision on the floor lasts 0.0830 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

Answer:

(a) $\vec{J}=176.4\frac{kg*m}{s} \hat{j}$

(b) $F=2125.30N$

Explanation:

(a) According to the law of conservation of energy, the potential energy of the person at 0.40 m is equal to its kinetic energy before the colision with the floor:

$\Delta U=\Delta K\\mgh=\frac{mv^2}{2}\\v=\sqrt{2gh}\\v=\sqrt{2(9.8\frac{m}{s^2})(0.40m)}\\v=2.8\frac{m}{s}$

This is the initial velocity in the negative y-direction. Impulse is given by:

$\vec{J}=\Delta \vec{p}\\\vec{J}=m\vec{v_f}-m\vec{v_i}\\\vec{J}=63kg(0 \hat{j})-63kg(-2.8\frac{m}{s} \hat{j})\\\vec{J}=176.4\frac{kg*m}{s} \hat{j}$

(b) The average force is:

$F=\frac{J}{\Delta t}\\F=\frac{176.4\frac{kg*m}{s}}{0.083s}\\F=2125.30N$