Question

A horizontal spring with a 0.500 kg mass attached to one end, and the other end is secured to a wall. If the mass pulled 6.00 cm to the right of the equilibrium position, held constant for a moment, and released, what is the velocity of the mass when it is 4.00 cm to the left of the equilibrium position

Answer 1

To solve this question I had to search in google the spring constant = 375 N/m!

The velocity of the mass is 1.22 m/s.

Step-by-step explanation:

To solve this question I had to search in google the spring constant = 375 N/m!

We can find the velocity of the mass by energy conservation:

`E_(i) = E_(f)`

`(1)/(2)kx_(i)^(2) = (1)/(2)mv^(2) + (1)/(2)kx_(f)^(2)`

Where:

m: is the mass = 0.500 kg

v: is the velocity of the mass=?

k: is the spring constant = 375 N/m (found in google)

`x_(i)` and
`x_(f)`: are the initial and final position of the spring respectively

`mv^(2) =k(x_(i)^(2) - x_(f)^(2))`

`v = \sqrt{(k(x_(i)^(2) - x_(f)^(2)))/(m)} = \sqrt{(375 N/m[(0.060m)^(2) - (-0.04 m)^(2)])/(0.500 kg)} = 1.22 m/s`

Therefore, the velocity of the mass is 1.22 m/s.

I hope it helps you!