Solving Work, Energy & Friction for Block Movement & Spring Compression

[joex444]

Just need a simple logic check here. Let's say a block of 2kg is positioned on an incline of 53 degrees. 4m from the block is a spring with a k=70N/m. the coefficient of kinetic friction is 0.36. How fast is the block moving when it hits the spring? How far is the spring compressed.

Ok, so find the height from the block to the spring (this is <4m), and say Ugrav = Wfriction + KE, to answer the first part. Then, that KE = Uspring + Wfriction to find the compression. Now, Wfriction would be Ff(x), where x is the compression of the spring. Yes? Or, would KE = Uspring be the correct way?

 

[Doc Al]

Ok, so find the height from the block to the spring (this is <4m), and say Ugrav = Wfriction + KE, to answer the first part.

Sounds good.

Then, that KE = Uspring + Wfriction to find the compression. Now, Wfriction would be Ff(x), where x is the compression of the spring. Yes? Or, would KE = Uspring be the correct way?

When solving for the compression, include the work done against friction. But don't forget the change in gravitational PE.

 

[andrewchang]

you may want to try solving the second part of this problem by skipping the kinetic energy- going directly from its original position to the final position. So, if x were the distance that the spring were compressed:

U + W_f = EPE
mg(h+x) + W_f = 0.5kx^2

 

[joex444]

Hmm...forgot about gravity after it hits the spring. Ok, so (H and X are the distance on the inclined plane, so I really want Hx and Xx in Ugrav)...
mg(h+x)\sin\theta + \mu mg\sin\theta = \frac{1}{2} kx^2
and I know m, g, h, theta, mu, and k, so 1 variable left x. I ended up with a quadratic, once I put numbers into it: 0=35x^2-15.65x-21.28 so x=1.03m. The question did say it was a "long" spring. I forgot gravity on the 2nd part on the final, so I ended up with an answer of 0.802m...

 

[mukundpa]

Wait; check the work by(or against) friction. The term does not appear dimensionally correct.

 

[Doc Al]

Hmm...forgot about gravity after it hits the spring. Ok, so (H and X are the distance on the inclined plane, so I really want Hx and Xx in Ugrav)...
mg(h+x)\sin\theta + \mu mg\sin\theta = \frac{1}{2} kx^2

Several problems with the 2nd term (the work done by friction):
(1) As mukundpa points out, it is dimensionally incorrect. (You forgot the distance.)
(2) The sign is incorrect.
(3) The sine is incorrect.