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# A movie stuntman (mass 80.0kg) stands on a window ledge 5.0 mabove the floor. Grabbing a rope attached to a chandelier, heswings down to grapple with the movie's villian (mass 70.0 kg), whois standing directly under the chandelier.(assume that thestuntman's center of mass moves downward 5.0 m. He releasesthe rope just as he reaches the villian). a) with what speed do the entwined foes start to slide acrossthe floor? b) if the coefficient of kinetic friction of their bodies withthe floor is 0.250, how far do they slide? # A movie stuntman (mass 80.0kg) stands on a window ledge 5.0 mabove the floor. Grabbing a rope attached to a chandelier, heswings down to grapple with the movie's villian (mass 70.0 kg), whois standing directly under the chandelier.(assume that thestuntman's center of mass moves downward 5.0 m. He releasesthe rope just as he reaches the villian). a) with what speed do the entwined foes start to slide acrossthe floor? b) if the coefficient of kinetic friction of their bodies withthe floor is 0.250, how far do they slide?

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Other asked 2021-05-13
A movie stuntman (mass 80.0kg) stands on a window ledge 5.0 mabove the floor. Grabbing a rope attached to a chandelier, heswings down to grapple with the movie's villian (mass 70.0 kg), whois standing directly under the chandelier.(assume that thestuntman's center of mass moves downward 5.0 m. He releasesthe rope just as he reaches the villian).
a) with what speed do the entwined foes start to slide acrossthe floor?
b) if the coefficient of kinetic friction of their bodies withthe floor is 0.250, how far do they slide?

## Answers (1) 2021-05-15
Given that
The mass of the stuntman is $$\displaystyle{\left({m}_{{{s}}}\right)}={80.0}{k}{g}$$
The mass of the movies villian is $$\displaystyle{\left({m}_{{{v}}}\right)}={70.0}{k}{g}$$
The distance between the window and floor is(y) =5.0m
The initial velocity is(u) =0
a)
By using the equation of motion we get that
The stuntman's speed before collision is $$\displaystyle{\left({v}_{{{0}}}\right)}=\sqrt{{{2}{g}{y}}}$$
The speed after collision is given by the conservation law
$$\displaystyle{v}={\left(\frac{{m}_{{{s}}}}{{m}_{{{s}}}}+{m}_{{{v}}}\right)}\cdot{v}_{{{0}}}$$
b)
The coefficient of kinetic friction of their bodies with thefloor is $$\displaystyle{\left(\mu_{{{k}}}\right)}={0.250}$$
During sliding momentum is not conserved.So, from the work-energy theorem we get that
Here x is the distance moved
$$\displaystyle\frac{{1}}{{2}}{m}_{{{t}}}{v}^{{{2}}}=\mu_{{{k}}}{m}_{{{t}}}{g}{h}$$
$$\displaystyle{x}=\frac{{v}^{{{2}}}}{{2}}\mu_{{{k}}}{g}$$
Now substitute all the values you get the required answer

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