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katrin2010 [14]

2 years ago

6

The spring of a spring balance is 6.0 in. long when there is no weight on the balance, and it is 8.3 in. long with 9.0 lb hung f

rom the balance. How much work is done in stretching it from 6.0 in. to a length of 12.2 in.

1 answer:

Aleks04 [339]2 years ago

5 0

Answer:

74.958 lb-in

Step-by-step explanation:

We are given that

Natural length of spring=l=6 in

After stretching, the length of spring=8.3 in

Weight=W=9 lb

Extension in the sprig=8.3-6=2.3 in

By hook's law

$F==W=kx$

Using the formula

$9=k(2.3)$

$k=\frac{9}{2.3}=3.9 lb/in$

Extension in the spring when the length of spring after stretching to 12.2 in=12.2-6=6.2 in

Work done, W= $\int_{0}^{6.2}3.9xdx$

$W=[\frac{3.9x^2}{2}]^{6.2}_{0}$

$W=3.9\times \frac{(6.2)^2}{2}$

$W=74.958 lb-in$

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