Solving Elasticity Question: Find Height of Block Above Release

[crazyeyes06]

A 0.230 kg block on a vertical spring with spring constant of 4.75 103 N/m is pushed downward, compressing the spring 0.058 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?

I have some equations in front of me but i don't know which one to use because it seems none of them have to do with height. Can anyone help me?

 

[Doc Al]

Try conservation of energy.

 

[m2003]

Based on the given information, we can use the principle of conservation of energy to solve for the height of the block above the point of release.

Initially, the block has potential energy stored in the compressed spring and no kinetic energy. At the point of release, the spring is no longer compressed and all of the potential energy is converted into kinetic energy.

Using the equation for potential energy, PE = (1/2)kx^2, where k is the spring constant and x is the displacement of the spring, we can calculate the initial potential energy of the block:

PE = (1/2)(4.75 x 10^3 N/m)(0.058 m)^2 = 8.337 J

At the point of release, all of this potential energy is converted into kinetic energy, which can be calculated using the equation KE = (1/2)mv^2, where m is the mass of the block and v is its velocity.

Setting the initial potential energy equal to the final kinetic energy, we can solve for the velocity of the block at the point of release:

8.337 J = (1/2)(0.230 kg)v^2

v = 7.253 m/s

Since the block is traveling vertically, we can use the equation for displacement to solve for the height:

y = y0 + v0t + (1/2)at^2

where y0 is the initial height, v0 is the initial velocity (which is 0), a is the acceleration due to gravity (9.8 m/s^2), and t is the time.

Solving for t and plugging in the known values, we get:

0.058 m = 0 + (1/2)(9.8 m/s^2)t^2

t = 0.34 s

Finally, plugging in the calculated time and initial velocity into the equation for displacement, we can solve for the height:

y = 0 + (0 m/s)(0.34 s) + (1/2)(9.8 m/s^2)(0.34 s)^2 = 0.56 m

Therefore, the block will rise 0.56 m above the point of release before it starts to fall back down due to gravity.