Question

A common practical joke is to pull a seat out from under someone as they attempt to sit down. The person falls to the floor and nobody has fun. Assume this happens to a 65 kg person, they fall 0.48 m, and it takes the floor 0.078 s to bring them to rest. (a) What was the impulse exerted on the person by the floor? (b) What was the average force exerted on the person by the floor?

Answer 1

Answer:

a)  I = 199.36 N s

,  b) F = 2555.9 N

Explanation:

a) For this problem we will use the relationship between momentum and momentum, let's start by finding the speed with which the person reaches the floor, using energy

Initial

    Em₀ = U = m h y

Final

    $Em_{f}$ = ½ m v²

    Em₀ = $Em_{f}$

    mg y = ½ m v²

    v = √ 2gy

    v = √ 2 9.8 0.48 m

    v = 3,067 m / s

Let's calculate the momentum

    I = Δp = m $v_{f}$ - m v₀

When it hits the floor it stops so its final speed is zero

    I = 0 - m v₀

    I = - 65 3,067

    I = - 199.36 N s

The momentum of the floor is of equal magnitude but opposite direction

    I = 199.36 N s

b) we write the other impulse relationship

    I = F t

   F = I / t

   F = 199.36 / 0.078

   F = 2555.9 N