A man starts at the bottom of a 35m tall tower, and walks to the top. The man ha

Question

A man starts at the bottom of a 35m tall tower, and walks to the top. The man has a mass of mm = 75kg, and he is carrying a steel ball with mass mb = 2.0kg.

a) How much work did the man do against gravity?

b) The man drops the ball into a thermally isolated container filled with water on the ground 35m below him. What is the speed of the ball just before it hits the water in the container?

c) If the water that the man drops the ball into has an initial temperature of Ti = 25C and a mass of mw = 6.0kg, what will be the temperature of the water after the vball falls in it? Assume the container does not move during the collision. Remember: W + Q = K + Ug + Etherm

Explanation / Answer

a)The work done by the man is:

W = M g h

M = mm + mb = 75 + 2 = 77 kg

W = 77 x 9.8 x 35 = 26411 J

Hence, W = 26411 J

b)from conservation of energy

KE = PE

1/2 mb v^2 = mb g h

v = sqrt (2 gh) = sqrt (2 x 9.8 x 35) = 26.19 m/s

Hence, v = 26.19 m/s

c)Ti = 25 deg C mw = 6 kg

We assume that the KE of the ball gets dissipated in the form of heat.

KE = 686 J

686 = 6 x 4186 x (Tf - Ti)

0.027 + Ti = Tf => Tf = 0.027 + 25 = 25.027 deg

Hence, Tf = 25.027 deg

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