Question

A spring of spring constant 200 N m-1 is
compressed by 2 cm. The potential energy stored
in spring is​

Answer 1

STEP BY STEP EXPLANATION

Answer :-

200 N/m

Given :-

• Spring constant (K) = 200 N/m
• Compressed(x) = 2cm

To find:-

• Elastic potential energy

Solution :-

We have formula to find the elastic potential energy that is

When the spring is compressed by an amount of x from its unstretched  position the elastic potential energy stored in spring is

U = 1/2 kx²

where,

• U = elastic potential energy
• k = spring constant
• x = compression

Substituting the values ,

U = 1/2 ×200×2

U = 200×2/2

U = 400/2

U = 200 N/m

So, the elastic potential energy is 200 N/m

Know more :-

• Units of elastic potential energy is N/m
• Elastic potential energy is always positive
• At natural state of spring , the elastic potential energy is zero

Answer 2

Hola ⚘⚘

Answer:

• 200 N/m

Explanation:

Given :

• A spring of spring constant 200 N m-1 is compressed by 2 cm.

Tofind:

• potential energy stored in spring

Calculation:

• U = ½ kx²

where,

• U = elastic pe
• k = spring constant
• x = compression

>> U = ½ * 200 * 2

>> U = 200*2/2

>> U = 400/2

>> U = 200 N/m

Hence, the elastic potential energy is 200 N/m