A force of 30 N is required to hold a spring that has been stretched from its natural length of 20 cm to a length of 35 cm. How
much work is done in stretching the spring from 35 cm to 40 cm
1 answer:
Answer:
0.25 J
Explanation:
Given that
Force on the spring, F = 30 N
Natural length of the spring, l1 = 20 cm = 0.2 m
Final length of the spring, l2 = 35 cm = 0.35 m
Extension of the spring, x = 0.35 - 0.2 = 0.15 m
The other extension is 40 - 35 cm = 5 cm = 0.05 m
Work done = ?
Considering Hooke's Law of Elasticity.
We're told that the spring stretches through 15 cm, and thereafter asked to find the work done in stretching it through 5 cm
The question is solved in the attachment below
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