A massless spring of spring constant k , has extension y and potential energy E . It is now stretched from y to 2y . The increase in its potential energy is

Question

A massless spring of spring constant k , has extension y and potential energy E . It is now stretched from y to 2y . The increase in its potential energy is (A)  3E  (B)  2E  (C)  E  (D)  4E

Tardigrade Admin · Accepted Answer

The potential energy of the spring is E=(1/2) ky2 dots(i) Now it is stretched from y to 2y, so its potential energy becomes E'=(1/2)k(2y)2=2ky2=4E (using (i)) ∴ The increase in its potential energy is Δ E=E' -E =4E-E=3E

Q. A massless spring of spring constant $k$ , has extension $y$ and potential energy $E$ . It is now stretched from $y$ to $2 y$ . The increase in its potential energy is

$3 E$

$2 E$

$E$

$4 E$

The potential energy of the spring is
$E = \frac{1}{2} k y^{2} \dots (i)$
Now it is stretched from $y$ to $2 y$ , so its potential energy becomes
$E^{'} = \frac{1}{2} k (2 y)^{2} = 2 k y^{2} = 4 E$ (using (i))
$∴$ The increase in its potential energy is
$Δ E = E^{'} - E$
$= 4 E - E = 3 E$

Q. A massless spring of spring constant k , has extension y and potential energy E . It is now stretched from y to 2y . The increase in its potential energy is

3E

2E

E

4E

The potential energy of the spring is E=21​ky2…(i) Now it is stretched from y to 2y, so its potential energy becomes E′=21​k(2y)2=2ky2=4E (using (i)) ∴ The increase in its potential energy is ΔE=E′−E =4E−E=3E

Q. A massless spring of spring constant $k$ , has extension $y$ and potential energy $E$ . It is now stretched from $y$ to $2 y$ . The increase in its potential energy is

$3 E$

$2 E$

$E$

$4 E$

The potential energy of the spring is
$E = \frac{1}{2} k y^{2} \dots (i)$
Now it is stretched from $y$ to $2 y$ , so its potential energy becomes
$E^{'} = \frac{1}{2} k (2 y)^{2} = 2 k y^{2} = 4 E$ (using (i))
$∴$ The increase in its potential energy is
$Δ E = E^{'} - E$
$= 4 E - E = 3 E$