A force of 12 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching i t from its natural length to 11 in. beyond its natural length?
1 answer:
Answer:
Work done in stretching the spring = 7.56 lb-ft.
Step-by-step explanation:
Normal length of the spring = 8 in or ft
= ft
If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in
= ft
Force applied to stretch the spring = 12 lb
By Hook's law ,
F = kx [where k is the spring constant and x = length by which the spring is stretched]
12 = k( )
k = 18
Work done (W) to stretch the spring by ft will be
W =
=
=
= 9( )²
= 7.56 lb-ft
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