Question

A force of 8 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching it from its natural length to 14 in. beyond its natural length? W = ft-lb

Answer 1

Answer:

The work done in stretching it from its natural length to 14 in. beyond its natural length is W=8.17 ft-lb.

Step-by-step explanation:

We know that a force of 8 lb is required to hold a spring stretched 8 in. beyond its natural length.

This let us calculate the spring constant k as:

$F=kx\\\\k=F/x=(8 \;lb)/(8\;in)=1\;lb/in$

We know that work is, in an scalar form, the product of force and distance.

The force F is equal to the spring constant multiplied by the distance from the natural length.

Then, as the force changes with the distance from the natural length, we have to calculate integrating:

$W=\int_0^{14}F\,dx=\int_0^{14}kx\,dx\\\\\\W=\dfrac{1}{2}kx^2\left|^{14}_0=\dfrac{1}{2}(1\,lb/in)(14 \,in)^2-\dfrac{1}{2}(1\,lb/in)(0 \,in)^2$

$W=98lb\cdot in-0lb-in\\\\W=98(lb\cdot in)\cdot \dfrac{1 \,ft}{12\,in}=8.17\;lb\cdot ft$