1. Solve the problem. A spring has a natural length of 20 in. A force of 1400 lb
ID: 2827738 • Letter: 1
Question
1. Solve the problem.
A spring has a natural length of 20 in. A force of 1400 lb stretches the spring to 30 in.
(a) Find the spring constat for this spring.
(b) How much work does it take to stretch the spring from its natural length to 30 in?
(c) How far beyond its natural length will a 350 lb force stretch the spring?
2. The base of a rectangular tank measures 8 ft by 16 ft. The tank is 18 ft tall, and its top is 10 ft below ground level. The tank is full of water weighing 62.4 lb/ft^3. How much work does it take to empty the tank by pumping the water to ground level? Do NOT use the gravitational constant g = 32, in your calculations. Give your answer to the nearest ft-lb.
Explanation / Answer
1.
(a)
F= kx ; where F= force, k=spring constant, and x=change in length of the spring.
=> 1400=k (30-20) => k = 140 lb/in
(b) Work= Fx = (1400)(30-20) = 14000 lb-in
(c) Force of 350 lb is applied. Let the spring be extended by 'x' ft.
So, we have, 350= (140)(x) => x= 2.5 in
2.
Volume of tank= length x breadth x height = 8x16x18= 2304 ft^3
Given, Density= 62.4 lb/ft^3
So Mass of water in the tank= Volume x Density= 143769.6 lb
Work required would be equal to the work done in displacing the centre of mass of the water to ground level.
Since the tank is completely filled, the centre of mass would be located at the middle, at a depth of ( 10 + 18/2) = 19ft.
Work done in order to pump up the water to ground level, against gravity, is therefore
(assuming acceleration due to gravity as 'g' ft/s^2)
W = mh = (143769.6)(19) = 2731622.4
Therefore, work done is approximately 2731622 ft-lb
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