A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched
Explanation:
Work, W = 2 J
Initial distance, [tex]x_{1}[/tex] = 30 cm
Final distance, = 42 cm
Force, F = 30 N
Stretched length, x = ?
We know,
W = 1/2 kΔx²
Δx = 42-30 cm = 12 cm = 0.12 m
2 J = 1/2 k X (0.12)²
k = 277.77 N/m
According to Hooke's law,
F = kx
30 N = 277.77 X x
x = 0.108 m
x = 10.8 cm
A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.
