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 67 kg, and the collision on the floor lasts 0.095 s. (a) What is the magnitude of the impulse acting on the victim from the floor during the collision?

Answer 1

Answer:

a).

The magnitude force impulse is

$F=656.6 kg*m/s$

the average force is

$F=656.6 kg*m/s$

Explanation:

Using the conservation energy the potential energy in the high is equal to the kinetic energy just before the collision so:

$E_K=E_p$

$\frac{1}{2}*m*v^2=m*g*h$

Notice the mass can be cancel as a factor so:

$v^2=2*g*h$

$v=\sqrt{2*g*h}=\sqrt{2*9.8m/s^2*0.50m}$

$v=9.8 m/s$

Now the impulse is determinate by:

$F=m*(v_f-(-v_i))$

$kg*F=m*v_i=67*9.8m/s$

$F=656.6 kg*m/s$

The average force is also

$F'=\frac{F}{t}=\frac{656.6}{0.095s}$

$F'=6911.57 N$