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

Question

A movie stuntman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor as in (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 4.0 m. He releases the rope just as he reaches the villain.) With what speed do the entwined foes start to slide across the floor? Express your answer in meters per second to two significant figures. If the coefficient of kinetic friction of their bodies with the floor is mu_k = 0.265, how far do they slide? Express your answer in meters to two significant figures.

Explanation / Answer

Here ,

let the speed at the floor is v

part A)

at the bottom

speed of hero , u = sqrt(2 * g * h)

u = sqrt(2 * 9.8 * 4)

u = 8.85 m/s

Now ,using conservation of momentum

80 * 8.85= (80 + 70) * v

v = 4.722 m/s

the initial speed of entwined across the floor is 4.72 m/s

part B)

Now , acceleration of bodies ,

a= -u * g

a = -0.265 * 9.8

a = - 2.6 m/s^2

Now, using third equation of motion

v^2 - u^2 = 2 * a * d

- 4.722^2 = = -2 * a 2.6 * d

d = 4.29 m

the distance travelled on the floor is 4.29 m

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