A clown of mass m climbs a ladder of height H and then jumps off and lands in a
ID: 1339210 • Letter: A
Question
A clown of mass m climbs a ladder of height H and then jumps off and lands in a tub of water. The bottom of the ladder just touches the surface of the water.
(a) What is the work done on the clown due to the Earth’s gravity when he climbs to the top of the ladder?
(b) As the clown climbs at a constant speed, the potential energy of the earth-clown system increases and the kinetic energy of the clown stays constant. Where does the energy come from?
(c) Consider the system consisting of the clown, the tub of water and the Earth. How does the energy of this system at the time the clown leaves the ladder compare to the energy of the system just before the clown hits the water? Ur Pd Pd Pd Pd Uranium at rest 2 Palladium at rest 2 Palladium very far apart, moving away
(d) When the clown comes to a stop in the water, the temperature of the water increases. Why?
(e) Some time later (the clown is still sitting in the water slightly dazed), the temperature of the water has returned to its initial temperature. How much heat has flowed out of the water during that time?
(f) Suppose m = 60 kg, H = 5 m, and the mass of the water is Mw = 1000 kg. What was the increase in temperature of the water when the clown landed in it? The specific heat of water is 4.2 J/(g·K).) What approximations did you make to reach an answer?
Explanation / Answer
a) W_gravity = -m*g*H
b) from clown.
c) both are same.
d) the kinetic energy of the clown is converted to heat.
water obsorbs ths heat.so, temearture of water increases.
e) energy flown out = m*g*H
f) heat energy gained = m*g*H
= 60*9.8*5
= 2940 J
now Apply, Q = Mw*C*dT
==> dT = Q/(Mw*C)
= 2940/(1000*4200)
= 7*10^-4 degrees celsius
We assume heat is not flowing from water to surroundings.
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