Question

If the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb, how much work is needed to stretch it 6 in. beyond its natural length?

Answer 1

Answer:

0.375 feet-lb

Step-by-step explanation:

We have been given that the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb. We are asked to find the work needed to stretch the spring 6 in. beyond its natural length.

We can represent our given information as:

$6=\int\limits^2_0 {F(x)} \, dx$

We will use Hooke's Law to solve our given problem.

$F(x)=kx$

Substituting this value in our integral, we will get:

$6=\int\limits^2_0 {kx} \, dx$

Using power rule, we will get:

$6=\left[ \frac{kx^2}{2} \right ]^2_0$

$6=\frac{k(2)^2}{2}-\frac{k(0)^2}{2}$

$6=\frac{4k}{2}-0\\\\k=3$

We know that 6 inches is equal to 0.5 feet.

Work needed to stretch it beyond 6 inches beyond its natural length would be $\int\limits^{0.5}_0 {kx} \, dx =\int\limits^{0.5}_0 {3x} \, dx$

Using power rule, we will get:

$\int\limits^{0.5}_0 {3x} \, dx = \left [\frac{3x^2}{2}\right]^{0.5}_0$

$\frac{3(0.5)^2}{2}-\frac{3(0)^2}{2}\Rightarrow \frac{3(0.25)}{2}-0=\frac{0.75}{2}=0.375$

Therefore, 0.375 feet-lb work is needed to stretch it 6 in. beyond its natural length.