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 0.50 m, the mass that moves downward is 70 kg, and the collision on the floor lasts 0.082 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) Impulse |J|= 219.4 kgm/s

b) Force F = 2672 N

Explanation:

Given

Height of fall h = 0.50 m

Mass M = 70 kg

Period of collision t = 0.082 s

Solution

The final velocity of the person v is zero since the person will come to rest.

The initial velocity of the person can be calculated by using the "law of conservation of energy".

Initial Kinetic energy = Final potential energy

$\frac{1}{2} mu^2=mgh\\\\u = \sqrt{2gh} \\\\u = \sqrt{2 \times 9.81 \times 0.50} \\\\u = 3.13 m/s$

a) Impulse

J = final momentum - initial momentum

$J = mv -mu\\\\J = 0 - (70 \times 3.13)\\\\J = -219.2 kgm/s$

Magnitude of impulse

$|J| = 219.1 kgm/s$

b) Force

$F = \frac{J}{t} \\\\F = \frac{219.1}{0.082} \\\\F = 2672 N$