Box and Spring Question Pt. II

[Iamaskier721]

Homework Statement

The green block is falling at a speed of 20 m/s and is 7 meters above the spring. The spring constant is 4155 N/m, to the nearest tenth of a cm how far is the spring compressed?

The answer to that part is 10.6cm

The next part is what I'm having difficulty with...

In the previous problem to the nearest hundredth of a meter to what height will the block rise after it hits and leaves the spring?

Homework Equations

V= (fx/m)^.5 which is the exity velocity I think. How will I find the maximum height?

 

[Hootenanny]

Homework Statement

The green block is falling at a speed of 20 m/s and is 7 meters above the spring. The spring constant is 4155 N/m, to the nearest tenth of a cm how far is the spring compressed?

The answer to that part is 10.6cm

The next part is what I'm having difficulty with...

In the previous problem to the nearest hundredth of a meter to what height will the block rise after it hits and leaves the spring?

Homework Equations

V= (fx/m)^.5 which is the exity velocity I think. How will I find the maximum height?

The question is actually a little simpler than you're making it out to be. If we assume that energy is conserved then the previous part of the question is irrelevant. All you need to know is that when the block is 7 meters above the spring it is traveling at 20m/s. What do you know about the velocity, and consequently the kinetic energy of the block at it's maximum height?

 

[baba89]

To find the maximum height, we can use the conservation of energy principle. At the moment the block hits the spring, it has a kinetic energy equal to 1/2mv^2, where m is the mass of the block and v is its velocity. This energy is then converted into potential energy as the spring compresses. When the block reaches its maximum height, all of its kinetic energy has been converted into potential energy. Therefore, we can set the initial kinetic energy equal to the final potential energy:

1/2mv^2 = mgh

Where g is the gravitational acceleration and h is the maximum height. We can rearrange this equation to solve for h:

h = v^2/2g

Plugging in the given values, we get:

h = (20 m/s)^2 / (2 * 9.8 m/s^2) = 20.41 m

To the nearest hundredth of a meter, the maximum height the block will reach is 20.41 m.