A 5.00-kg air-filled, sealed, rigid float tank that has a volume V = 1.40m^3 is
ID: 1291582 • Letter: A
Question
A 5.00-kg air-filled, sealed, rigid float tank that has a volume V = 1.40m^3 is pulled 50.4 m down to the seafloor in order to assist in lifting a sunken object. A diver standing on the seafloor cranks a winch to pull the tank down (Figure 1) .
Figure 1
Part A
How much work is required to pull the tank down? (Ignore the mass of the air in the tank and the mass of the winch rope. Assume that the water mass density does not change with depth.)
Express your answer to two significant digits and include the appropriate units.
W =
Part B
What is the average power required if it takes 10.0 min to winch down the tank?
Express your answer to two significant digits and include the appropriate units.
P =
A 5.00-kg air-filled, sealed, rigid float tank that has a volume V = 1.40m^3 is pulled 50.4 m down to the seafloor in order to assist in lifting a sunken object. A diver standing on the seafloor cranks a winch to pull the tank down (Figure 1) . Figure 1 Part A How much work is required to pull the tank down? (Ignore the mass of the air in the tank and the mass of the winch rope. Assume that the water mass density does not change with depth.) Express your answer to two significant digits and include the appropriate units. W = Part B What is the average power required if it takes 10.0 min to winch down the tank? Express your answer to two significant digits and include the appropriate units. P =Explanation / Answer
two forces act on the tank i.e. its weight(downward) and boyant force(upward)
W = mg = 5 * 9.8 = 49 N
Boyant force = weight of water displaced by the tank (Archemedes principle)
= Density of water * Volume * g
= 1000 kg/m3 * 1.40 m3 * 9.8 m/s2
= 13720 N
Net force = 13720 - 49 = 13671 N (downwaed)
So the force applied by winch = 13671 N
PART A
Work done = F.S =13671 *50.4 =689018.4 J = 6.9 *105 N
PART B
Power = work done /Time
P = 6.9 *105/ 10 *60 =1150 = 1.2 *103 W
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