Question

A force of 7 lb is required to hold a spring stretched 9 in. beyond its natural length. How much work W is done in stretching it from its natural length to 13 in. beyond its natural length

Answer 1

144 joules is the work W done in stretching it from its natural length to 13 in beyond its natural length

Force F=10 lb

From the hookes Law

$F=kx$

Therefore, calculate k for the spring for

$10=k(\frac{4}{12}) = > k=\frac{120}{4}=30$

Work done in stretching a spring through a length dx is

$dW=F.dx = > dW=kxdx$

For calculating the work done for stretching the spring to x=6in=(6/12)feet beyond the natural length, Integrate over the limits of x=0 to x=1/2

Therefore,

$\int_{0}^{W}dW=\int_{0}^{0.5}kxdx = > W=\frac{1}{2}k[x^2]_{0}^{0.5}=\frac{1}{2}30 \times (0.5)^2=\frac{30}{8}$

That is,

$W=\frac{15}{4} ft-lb$

The answer that is given is for stretching the spring to 4 inches.

Mass of 10 m of chain length is 80kg. This implies, Mass per unit meter length of the chain is

$m_{l}=\frac{80}{10}=8kg/m$

Consider a small length dx of the chain at the end point A. Work done in lifting the small length dx over the height of x meters is

$dW=(8g)dx \times x=8gxdx$

Now, integrate over the the value of x from

x=0 when end of the chain A is on the ground

x=6 when end of the chain A reaches 6 m above the ground

That is,

$\int_{0}^{W}dW=\int_{0}^{6}8gxdx=8g [\frac{x^2}{2}]_{0}^{6}=4g(6^2-0^2)$

$W=4g \times 36=144g \ Joules$

Hence,144 joules is the work W done in stretching it from its natural length to 13 in beyond its natural length

Learn more about Integration here brainly.com/question/2263647

#SPJ4