A movie stuntman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor (Fig. P8.81). Grabbing a rope attached to a chandelier, he swings down to grapple with the movie’s villain (mass 70.0 kg), who is standing directly under the chandelier. (Assume that the stuntman’s center of mass moves downward 5.0 m. He releases the rope just as he reaches the villain.) (a) With what speed do the entwined foes start to slide across the floor? (b) If the coefficient of kinetic friction of their bodies with the floor is , how far do they slide?

The height of the window ledge above the floor is h = 5 m

The mass of stuntmen is: M = 80 kg

The mass of villains is: m = 70 kg

The coefficient of kinetic friction between bodies and the floor is ${\mu }_k=0.250$.

The momentum conservation gives the speed of entwined foes and the work-energy theorem gives sliding distance on the floor.

The speed of the stuntman is given as:

$u=\sqrt{2gh}$

Here, g is the gravitational acceleration.

Substitute all the values in the above equation.

$$\begin{gathered} u=\sqrt{2\left(9.8m/s^2\right)\left(5m\right)} \\ u=9.9m/s \end{gathered}$$

Apply the momentum conservation to find the speed of entwined woes.

$Mu=\left(m+M\right)v$

Substitute all the values in the above equation.

$$\begin{gathered} \left(80kg\right)\left(9.9m/s\right)=\left(80kg+70kg\right)v \\ v=5.3m/s \end{gathered}$$

Therefore, the speed of entwined woes on floor is 5.3 m/s .

Apply the work energy theorem to find slide distance.

$$\begin{gathered} \frac{1}{2}\left(m+M\right)v^2={\mu }_k\left(m+M\right)gd \\ d=\frac{v^2}{2{\mu }_{k}g} \end{gathered}$$

Substitute all the values in the above equation.

$$\begin{gathered} d=\frac{{\left(5.3m/s\right)}^2}{2\left(0.250\right)\left(9.8m/s^2\right)} \\ d=5.73m \end{gathered}$$

Therefore, the sliding distance for entwined woes is 5.7 m .