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

Question

Physics

Cervantes

3 December, 10:15

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

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Answers (2)

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Leland · 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.

Peppy · 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