The work to stretch a spring from its rest position is
(1/2) (spring constant) (distance of the stretch)²
E = 1/2 k x² .
You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write
1700 joules = 1/2 k (3m)²
1 joule = 1 newton-meter
1700 N-m = 1/2 k (3m)²
Multiply each side by 2: 3400 N-m = k · 9m²
Divide each side by 9m² k = 3400 N-m / 9m²
= (377 and 7/9) newton per meter
The answer is distressing
The spring has a spring constant of 1.00 * 10^3 N/m and the mass has been displaced 20.0 cm then the restoring force is 20000 N/m.
Explanation:
When a spring is stretched or compressed its length changes by an amount x from its equilibrium length then the restoring force is exerted.
spring constant is k = 1.00 * 10^3 N/m
mass is x = 20.0 cm
According to Hooke's law, To find restoring force,
F = - kx
= - 1.00 *10 ^3 * 20.0
F = 20000 N/m
Thus, the spring has a spring constant of 1.00 * 10^3 N/m and the mass has been displaced 20.0 cm then the restoring force is 20000 N/m.
