A force of 20 pounds stretches a spring 11 inches in an exercise machine. Find the work done in stretching the spring 2 feet from its natural position.
A) $\frac { 240 } { 11 } \mathrm { ft } - \mathrm { lb }$
B) $\frac { 960 } { 11 } \mathrm { ft } - \mathrm { lb }$
C) $\frac { 66 } { 5 } \mathrm { ft } - \mathrm { lb }$
D) $\frac { 480 } { 11 } \mathrm { ft } - \mathrm { lb }$
E) $\frac { 33 } { 5 } \mathrm { ft } - \mathrm { lb }$

Question

A force of 20 pounds stretches a spring 11 inches in an exercise machine. Find the work done in stretching the spring 2 feet from its natural position. 
A) $\frac { 240 } { 11 } \mathrm { ft } - \mathrm { lb }$
B) $\frac { 960 } { 11 } \mathrm { ft } - \mathrm { lb }$
C) $\frac { 66 } { 5 } \mathrm { ft } - \mathrm { lb }$
D) $\frac { 480 } { 11 } \mathrm { ft } - \mathrm { lb }$
E) $\frac { 33 } { 5 } \mathrm { ft } - \mathrm { lb }$

Quizplus · Accepted Answer

We can use the formula for potential energy stored in a spring: 
$$U=\frac{1}{2}kx^2$$ 
where $U$ is the potential energy stored in the spring, $k$ is the spring constant, and $x$ is the displacement from the natural position.

To find the spring constant, we can use the given information: 
$$F=kx\Rightarrow k=\frac{F}{x}=\frac{20}{11}	ext{ lb/in}$$

If we want to stretch the spring 2 feet, or 24 inches, from its natural position, then the displacement is: 
$$x=24	ext{ in}-0	ext{ in}=24	ext{ in}$$

Plugging in the values, we get: 
$$U=\frac{1}{2}\left(\frac{20}{11}	ext{ lb/in}ight)\left(24	ext{ in}ight)^2=\frac{480}{11}	ext{ in-lb}=\frac{480}{11}\cdot\frac{1}{12}	ext{ ft-lb}=\frac{40}{11}	ext{ ft-lb}$$

Therefore, the work done in stretching the spring 2 feet is $\frac{480}{11}$ ft-lb, which is choice (D).

A Force of 20 Pounds Stretches a Spring 11 Inches 24011ft−lb\frac { 240 } { 11 } \mathrm { ft } - \mathrm { lb }11240​ft−lb

A Force of 20 Pounds Stretches a Spring 11 Inches $\frac { 240 } { 11 } \mathrm { ft } - \mathrm { lb }$