A weight of 200 N is hung from a spring with a spring constant of 2500 N/m and lowered slowly. How much will the spring stretch?
A) 4.00 cm
B) 6.00 cm
C) 8.00 cm
D) 10.0 cm
E) 12.0 cm

Question

Quizplus · Accepted Answer

We can use Hooke's law to find the amount of stretch in the spring: 
F = kx 
where F is the force (weight) applied, k is the spring constant, and x is the displacement (stretch) of the spring. Rearranging this equation, we get: 
x = F/k

Plugging in the given values, we get: 
x = 200 N / 2500 N/m 
x = 0.08 m 
x = 8.00 cm

Therefore, the spring will stretch by 8.00 cm.

A Weight of 200 N Is Hung from a Spring