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

Answer:

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.

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:

$E_{i} = E_{f}$

$\frac{1}{2}kx_{i}^{2} = \frac{1}{2}mv^{2} + \frac{1}{2}kx_{f}^{2}$

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

$mv^{2} =k(x_{i}^{2} - x_{f}^{2})$

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

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

                                                                       

I hope it helps you!