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 0.60 m, the mass that moves downward is 79.0 kg, and the collision on the floor lasts 0.0840 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) I = 270.18 Kg*m/s

b) F = -3216.42N

Explanation:

a) We know that:

I = $P_f -P_i$

Where I is the impulse, $P_f$ is the final momentum and $P_i$ the initial momentum.

so:

I = $MV_f -MV_i$

where M is the mass, $V_f$ is the final velocity and $V_i$ is the initial velocity.

First we have to find the $V_i$. So, using the conservation of energy.

$Mgh = \frac{1}{2}MV_i^2$

where g the gravity and h the altitude. Replacing values, we get:

$(79kg)(9.8m/s)(0.6m) = \frac{1}{2}(79kg)V_i^2$

solving for $V_i$:

$V_i$= 3.42 m/s

Now, replacing in the previus equation:

I = $MV_f -MV_i$

I = $(79kg)(0)-(79kg)(3.42m/s)$

I = -270.18 Kg*m/s

The impulse is negative becuase it is upward.

b) We know that:

Ft = I

where F is the force, t the time and I the impulse.

so, replacing values and solving for F, we get:

F(0.084s) = -270.18 Kg*m/s

F = -3216.42N

The force is negative becuase it is upward.