Question

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 8.50 N is applied. A 0.600-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x

Answer 1

Answer:

Spring constant, k = 283.33 N/m

Explanation:

Given that,

Force acting on the spring, F = 8.5 N

Stretching in the spring, x = 3 cm = 0.03 m

Let k is the spring constant of the spring. It can be calculated using Hooke's law as :

$F=-kx$

$k=\dfrac{F}{x}$

$k=\dfrac{8.5\ N}{0.03\ m}$

k = 283.33 N/m

So, the spring constant of the spring is 283.33 N/m. Hence, this is the required solution.