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A force of 10 lb is required to hold a spring stretched 2 in. beyond its natural length. How much work W is done in stretching it from its natural length to 5 in. beyond its natural length?

Respuesta :

Answer:

Work done will be equal to 5.2059 lb-ft

Step-by-step explanation:

We have given force F = 10 lb

Spring is stretched to 2 in

So x = 2 in

As 1 inch = 0.0833 feet

So 2 inch = 2×0.0833 = 0.1666 feet

From hook's law we know that F = Kx , here K is spring constant and x is spring elongation

So [tex]10=K\times 0.1666[/tex]

K = 60.024 lb/feet

Now new elongation x = 5 in

So 5 in = 5×0.0833 = 0.4165 feet

Work done is given by [tex]W=\frac{1}{2}Kx^2[/tex]

So [tex]W=\frac{1}{2}\times 60.02\times 0.4165^2=5.205lb-ft[/tex]

So work done will be equal to 5.2059 lb-ft

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