A movie stuntman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor (

Question

A movie stuntman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor (Figure 1) . 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.)

With what speed do the entwined foes start to slide across the floor?

If the coefficient of kinetic friction of their bodies with the floor is ?k = 0.215, how far do they slide?

A movie stuntman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor (Figure 1) . 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.) With what speed do the entwined foes start to slide across the floor? If the coefficient of kinetic friction of their bodies with the floor is ?k = 0.215, how far do they slide?

Explanation / Answer

Stuntman falls 5m, so equate PE with KE.

A)

mgh = (1/2)mv

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