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.
