# Need to find the distance for the physics problem.

1. Oct 12, 2012

### NasuSama

1. The problem statement, all variables and given/known data

An incline makes an angle of 21.2o with the horizontal. A 4.53 kg block is given a push up this incline and released. It starts at the bottom with initial speed 2.71 m/s, travels up the incline, stops, and slides back to the bottom at final speed 1.82 m/s. Using energy considerations, find:

the distance the block traveled up along the incline before coming momentarily to rest.

2. Relevant equations

Hm....

→½mv²
→mgh
→trig identity

3. The attempt at a solution

½mv_i² = mgh
½v_i² = gh
½ * v_i² / g = h

I believe h = sin(θ)d, so...

½ * v_i² / g = sin(θ)d
d = ½ * v_i² / (g * sin(θ))

But the approach is incorrect

Last edited: Oct 12, 2012
2. Oct 12, 2012

### HallsofIvy

Staff Emeritus
??? "Using energy considerations, find:" Find what? You give some formulas but don't apply them to the numbers? What are you trying to find and why do you say "the approach is incorrect"?

3. Oct 12, 2012

### NasuSama

Sorry about this. I want to find the distance the block traveled up along the incline before coming momentarily to rest.

4. Oct 12, 2012

### NasuSama

5. Oct 12, 2012

### PhanthomJay

Given that the block has a different speed than its original speed when it returns to its start point, should that tell you something about what forces might be acting to cause this?

6. Oct 12, 2012

### NasuSama

Is it energy used to overcome friction, which I found is 9.13 J?

Yes, it's frictional force, but I need to only use potential and kinetic energy formulas (or energy considerations)

7. Oct 12, 2012

### NasuSama

Then, this means that I need to use the energy used to overcome the frictional force? Does this equation work?

W_f = mgdsin(θ)µ_k

I am not sure if the mgh thing is important to consider.

* Edit * Nevermind. That is not even right.

O.K. Seems like there is no right approach for me. -__-

Last edited: Oct 12, 2012
8. Oct 12, 2012

### NasuSama

I actually got the solution by Newton's Law.

9. Oct 13, 2012

### PhanthomJay

Newton 2 is always a good backup, although it involves a bit more steps than the energy approach you were asked to use. You already have calculated correctly the energy loss from friction between the start point at the bottom of the incline and end point at the bottom of the incline (same point). That represents the work done by friction. Half of that for the up motion and half for the down motion. Use the up part in your work - energy equation..work done by friction is the sum of the changes in potential and kinetic energies.

10. Oct 13, 2012

### NasuSama

That is actually the approach by energy considerations. Thanks!