Question

A crate is hung from a spring with a force constant of 525 nm. this stretches the spring 0.30 m from its equilibrium position. what is the mass of the crate? use 10 m/s2

Answer 1

16 kg.
Hooke's law is:
F = kX
where
F = Force
X = displacement of spring
k = force constant for spring.
So let's calculate F
F = kX
F = 525 N/m * 0.30m
F = 157.5 N
F = 157.5 kg*m/s^2

Now that we know the force, we can calculate the mass by dividing by the gravitational acceleration which has been given to use as 10 m/s^2. So 157.5 kg*m/s^2 / 10 m/s^2 = 15.75 kg

So the mass of the crate is 15.75 kg. Rounding to 2 significant figures gives 16 kg.

Answer 2

Lets use the expression

F = k*x

Where k is the spring constant [N/m]

And x the distance from the equilibrium position.

This force is equal to the force due to the acceleration of gravity

F = k*x = m*g

F = 525[N/m]* 0.30 [m] = 157.5 [N]

157.5 [N] = m *10[m/s**2] ...................> m = 15.75 kg