Question

A spring has a natural length of 8 m. If a 12-N force is required to keep it stretched to a length of 10 m, how much work W is required to stretch it from 8 m to 16 m? (Round your answer to two decimal places.)

Answer 1

Answer:

The work is required to stretch it from 8 m to 16 m is 192 N-m

Explanation:

Given that,

Natural length = 8 m

Force F = 12 N

After stretched,

length = 10 m

We need to calculate the elongation

$x = 10-8=2\ m$

Using hook's law

The restoring force is directly proportional to the displacement.

$F\propto (-x)$

$F = -kx$

Where, k = spring constant

Negative sign shows the displacement in opposite direction

Now, The value of k is

$k = \dfrac{F}{x}$

$k = \dfrac{12}{2}$

$k = 6$

When stretch the string from 8 m to 16 m.

Then the elongation is

$x=16-8=8\ m$

Now, The work is required to stretch it from 8 m to 16 m

$W = \dfrac{1}{2}kx^2$

Where, k = spring constant

x = elongation

$W=\dfrac{1}{2}\times6\times8\times8$

$W=192\ N-m$

Hence, The work is required to stretch it from 8 m to 16 m is 192 N-m