Question

a force of 5 N extends a spring of natural length 0.5m by 0.01, what will be the length of the spring when the applied force is 20 N​

Answer 1

Answer:

L_new =L+x^2  = L_new = 0.54_m.

Explanation:

Given data:

Force in the first case,  

F_1 = 5N

Force in the second case,  

F_2 = 20 N

Natural length of spring,  

L= 0.5

Extension in the first case,  

x_1 = 0.01m

Let the force constant of the spring be k.

Thus,

F_1=kx_1

5 = k × 0.01  

⇒ k = 500 N/m.

The extension in the spring in the second case can be given as,

F_2=kx_2

20 = 500x_2

⇒ x_2 = 0.04 m.

Thus, the effective length of the spring would be,

L_new =L+x^2

L_new = 0.5+0.04

L_new = 0.54_m.