- #1

spacestrudel

- 12

- 0

- Homework Statement
- A block is sliding down an inclined plane with friction. The block collides with a spring at the bottom of the incline. What is the max. compression of the spring?

- Relevant Equations
- Spring Potential Energy = 1/2kx^2. Ei = Ef.

Hello! I am stuck on part of a problem and was wondering what I am doing wrong. For part a of the problem, we were asked to find the impact speed. I did this in a photo below given the following values:

Θ = 30 degrees. The initial velocity = 10 m/s. The coefficient of kinetic friction = 0.4. distance between the spring and block = 5 meters. Mass of the block = 300kg. k = 20N/m.

My professor told us that when the spring is at max. compression, the velocity equals zero. So, I'm assuming we use that somehow?

Here is a photo of my work thus far (sorry it's blurry on here):

Initially I tried to perform the following calculations:

mg(hf + x) = 1/2kx^2

= 1/2kx^2 = mg(x+5)sin(30)

= (300)(9.81)(x+5)(sin(30)) = 100x^2

= (2,943 N)(x+5)(sin(30)) = 100x^2

Which eventually simplifies to the following:

(100)x^2 - (1471.5)x - 7357.5 = 0.

I plugged these values into the quadratic formula and received x = 18.658 meters. However, this seems far too large to be the maximum compression considering there is friction and I'm not sure how to take that into account. Should I perhaps multiply the friction times sin(30)? Or maybe just scrap this method and try a different set of equations?

Thanks so much!

Θ = 30 degrees. The initial velocity = 10 m/s. The coefficient of kinetic friction = 0.4. distance between the spring and block = 5 meters. Mass of the block = 300kg. k = 20N/m.

My professor told us that when the spring is at max. compression, the velocity equals zero. So, I'm assuming we use that somehow?

Here is a photo of my work thus far (sorry it's blurry on here):

Initially I tried to perform the following calculations:

mg(hf + x) = 1/2kx^2

= 1/2kx^2 = mg(x+5)sin(30)

= (300)(9.81)(x+5)(sin(30)) = 100x^2

= (2,943 N)(x+5)(sin(30)) = 100x^2

Which eventually simplifies to the following:

(100)x^2 - (1471.5)x - 7357.5 = 0.

I plugged these values into the quadratic formula and received x = 18.658 meters. However, this seems far too large to be the maximum compression considering there is friction and I'm not sure how to take that into account. Should I perhaps multiply the friction times sin(30)? Or maybe just scrap this method and try a different set of equations?

Thanks so much!