A force of 10 lb is required to hold a spring stretched 2 in. beyond its natural

Question

A force of 10 lb is required to hold a spring stretched 2 in. beyond its natural length. How much work W is done in stretching it from its natural length to 5 in. beyond its natural length? W =_______ft-lb A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 for g and 3.14 for pi. If you enter your answer in scientific notation, round the decimal value to two decimal places. Use equivalent rounding if you do not enter your answer in scientific notation.) w =_________J r = 6 h = 1.5 A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 for g.) W =__________J a = 4 b = 4 c = 15 d = 2

Explanation / Answer

A)

f= 10 lb for stretch of 2 inch

for strching of 5 inches force required 25lb( 10*5/2)

work done = integral of fdx in interval 0 to 5

w = 0 to 5 integral fdx = (fx)05 is 25(5-0)= 125 lb-inch

part b)

w = ((4/3)(pi)r^3)g/h

= [(4/3)(3.14)(6)^3](9.8/1.5)

= 5908.224 J

part c)

a= 4, b=4,c=15,d=2

volume = (.5)*a*b*c = (.5)*4*4*15 = 120

work = 120(2 - 0) = 240 J

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