Question

the potential energy of a spring increases by 15 J when stretched by 3 cm . If it is stretched by 4 cm, the increase in potential energy is​

Answer 1

The potential energy of a spring increases by 15 J when stretched by 3 cm. If it is stretched by 4 cm, the increase in potential energy is​ 26.64J.

Given:

$d_{1} = 3cm$

$d_{2} = 4cm$

P.E = 15J

To find:

P.E when stretched by 4cm = ?

Solution:

Potential energy:

Whenever we extend or compress any spring, a force is experienced which tries to get the spring back in original position, this experienced energy is spring potential energy.

Formula to be used:

P.E = $\frac{1}{2} kx^{2}$

$15 = \frac{1}{2} k (3)^{2}\\\\30 = k(3)^{2} \\\\30 = 9k\\\\k = 3.33$

Since, we got k = 3.33

Substituting value of k in equation for 4cm,

$P.E = \frac{1}{2}(3.33)(4)^{2} \\ \\P.E = 26.64J$

Therefore, if it is stretched by 4 cm, the increase in potential energy is​ 26.64J