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qwelly [4]
3 years ago
9

The spring of a spring balance is 6.0 in. long when there is no weight on the balance, and it is 8.4 in. long with 4.0 lb hung f

rom the balance. How much work is done in stretching it from 6.0 in. to a length of 10.1 in.?
Physics
1 answer:
Sergeeva-Olga [200]3 years ago
3 0

Answer:

W = 55.12 J

Explanation:

Given,

Natural length = 6 in

Force = 4 lb,  stretched length = 8.4 in

We know,

F = k x

k is spring constant

4 = k (8.4-6)

k = 1.67 lb/in

Work done to stretch the spring to 10.1 in.

W =k\int_{6}^{10.1} x

W = \dfrac{k}{2}[x^2]_6^{10.1}

W = \dfrac{1}{2}\times 1.67\times (10.1^2-6.0^2)

W = 55.12 J

Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.

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