Answer:
8.4 ft-lb
Explanation:
Work = change in energy
W = ½ kx²
When x = 3 ft, W = 15 ft-lb:
15 ft-lb = ½ k (3 ft)²
k = 30/9 lb/ft
When x = 27 in = 2.25 ft:
W = ½ kx²
W = ½ (30/9 lb/ft) (2.25 ft)²
W = 8.4375 ft-lb
Rounding to 2 sig-figs, it takes 8.4 ft-lb of work.
The work needed to stretch it 27 in. beyond its natural length is 8.4 ft-lb
Work done by an object is equal to the potential energy of the spring.
W = ½ kx²
Where, displacement x = 3 ft, Work done W = 15 ft-lb:
Substituting the values, we get the spring constant
15 ft-lb = ½ k (3 ft)²
k = 30/9 lb/ft
When the displacement x = 27 in = 2.25 ft, work done will be
W = ½ kx²
W = ½ (30/9 lb/ft) (2.25 ft)²
W = 8.4375 ft-lb
Thus, the work needed to stretch is approximately 8.4 ft-lb
Learn more about work done.
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