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 69 kg, and the collision on the floor lasts 0.087 s.

(a) What is the magnitude of the impulse acting on the victim from the floor during the collision? N·s
(b) What is the magnitude of the average force acting on the victim from the floor during the collision?

Answer 1

Step-by-step explanation:

Answer:

Step-by-step explanation:

mass of person, m = 69 kg

distance, h = 0.5 m

time, t = 0.087 s

(a)

Let he strikes with the ground with velocity u

By using the conservation of energy

m g h = 1/2 mu²

9.8 x 0.5 = 0.5 x u²

u = 3.13 m/s

(a) Impulse = Change in momentum

I = m u - m v

As the final velocity is zero, v = 0

I = 69 x 3.13 = 215.97 N s

(b)

Average force = Impulse / time = 215.97 / 0.087 = 2482.4 N