Question

A steel spring of spring constant 5N/m is initially 10 cm long. Calculate the increase in its elastic potential energy when its length is increased to 15 cm

Answer 1

Answer:-

• 6.25 × 10-³ J

Step-by-step explanation:

The elastic potential energy of a spring is given as ,

`\implies\boxed{ \rm{ Elastic\ PE }\ = (1)/(2) kx^2 } \\\\`

where,

• 
  `x` is the displacement from the mean position.
• 
  `k` is spring constant.

Change in Elastic potential energy of a spring is given as,

`\implies\boxed{ \Delta \rm{ Elastic\ PE \ }= (1)/(2)(x_2-x_1)^2 }\\\\`

where,

• 
  `x_1` is the initial position .
• 
  `x_2` is the final position.
• 
  `k` is spring constant.

On substituting the respective values, we have;

`\implies \Delta \rm{Elastic\ PE } \ = (1)/(2) 5\ N/m ( 15cm - 10cm )^2 \\\\`

`\implies \Delta \rm{Elastic\ PE } \ = (1)/(2)* 5\ N/m (0.05m)^2 \\\\`

`\implies \Delta \rm{Elastic\ PE } \ = 2.5 \ N/m * (25)/(10^4) \\\\`

`\implies \Delta \rm{Elastic\ PE } \ = 0.00625 \ J \\\\`

`\implies \boxed{\Delta \rm{Elastic\ PE } \ = 6.25 * 10^(-3) \ J }\\\\`

Henceforth the increase in elastic potential energy is 6.25 × 10-³ J .

Answer 2

The increase in the elastic potential energy of a steel spring, with a spring constant of 5 N/m, when its length is increased from 10 cm to 15 cm is 0.625 J.

Step-by-step explanation:

Given:

Spring constant (k) = 5 N/mInitial length of spring (li) = 10 cmFinal length of spring (lf) = 15 cm

To calculate the increase in elastic potential energy, we need to find the difference in lengths of the spring (Δl) and use the formula for elastic potential energy:
Δl = lf - li = 15 cm - 10 cm = 5 cm
Elastic potential energy (PE) = 1/2 * k * (Δl)2
Substituting the given values,
PE = 1/2 * 5 N/m * (5 cm)2

Simplifying the equation,

PE = 1/2 * 5 N/m * 25 cm2 = 62.5 N·cm = 0.625 J

Therefore, the increase in the spring's elastic potential energy when its length is increased from 10 cm to 15 cm is 0.625 J.