Question

A spring of force constant K is stretched a certain distance.it takes twice as much work to stretch a second spring by half this distance.the force constant of the second spring is

Answer 1

Force constant of first spring $\sf{=k}$

Let the force constant of second spring be $\sf{k_2.}$

Work done on stretching the first spring by a certain distance $\sf{x}$ (say) is,

$\longrightarrow\sf{W=\dfrac{1}{2}\,kx^2}$

Work done on stretching the second spring by half the distance by which first spring is stretched, i.e., $\sf{\dfrac{x}{2},}$ is,

$\longrightarrow\sf{W_2=\dfrac{1}{2}\,k_2\left(\dfrac{x}{2}\right)^2}$

$\longrightarrow\sf{W_2=\dfrac{1}{8}\,k_2x^2}$

Given that the work done on first spring is as twice as the work done on second one, i.e.,

$\longrightarrow\sf{W=2W_2}$

$\longrightarrow\sf{\dfrac{1}{2}\,kx^2=2\times\dfrac{1}{8}\,k_2x^2}$

$\longrightarrow\sf{\dfrac{1}{2}\,k=\dfrac{1}{4}\,k_2}$

$\longrightarrow\sf{\underline{\underline{k_2=2k}}}$

Hence $\bf{2k}$ is the answer.

Answer 2

Answer:

8k

Explanation:

Force constant of first spring

k1

Let the force constant of second spring be

k2

Work done on stretching the first spring by a certain distance  (say) is,

w1 = 1/2 k1 x^2 (1)

Work done on stretching the second spring by half the distance by which first spring is stretched, i.e.,  is,

w* = 1/2 k2 (1/2 x) ^2=1/4 (1/2 k2 x^2)

Given that the work done on second spring is as twice as the work done on first one, i.e.,

w2 = 2 w*  

2 w* =  (1/4 (1/2 k2 x^2)  (2)

by dividing (1) / (2)

1/2= 4 k1/k2

k2 = 8 k1

Hence 8k is the answer.