Question

if stretch in a spring of force constant k is doubled , calculate (a)the ratio of final to initial force in the spring (b)the ratio of elastic energies stored in two cases.​

Answer 1

Answer:

2:1 will be the answers of this question not the above mention.

Answer 2

Let the spring force constant be k

Let the new spring force constant be k'

Since the spring force constant is doubled, k' = 2k

Force due to spring is given as

F = kx, where F is the force, k is the spring force constant and x is the extension in the spring

Initial force (F) = kx

Final force (F') = k'x

                        = 2kx

(a) $\frac{Final force(F')}{Initial force(F)}$ = $\frac{k'x}{kx}$

$\frac{Final force(F')}{Initial force(F)}$ = $\frac{2kx}{kx}$

$\frac{Final force(F')}{Initial force(F)}$ = $\frac{2}{1}$

(b) Elastic energy stored in the spring is given as

E = $\frac{1}{2}$kx², where E is the elastic energy stored, k is the spring force constant, x is the extension in the spring

Initial Elastic energy (E) = $\frac{1}{2}$kx²

Final elastic energy (E') =  $\frac{1}{2}$k'x²

                                       = $\frac{1}{2}$ 2kx²

                                       =  kx²

$\frac{E}{E'}$ = $\frac{\frac{1}{2}kx^{2} }{kx^{2} }$

$\frac{E}{E'}$ = $\frac{1}{2}$  

Therefore, the ratio of final force to initial force is 2:1 and the ratio of initial elastic energy to final elastic energy is 1:2