Need to find the distance for the physics problem.

[NasuSama]

Homework Statement

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.

Homework Equations

Hm...

→½mv²
→mgh
→trig identity

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

 

[HallsofIvy]

? "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"?

 

[NasuSama]

? "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"?

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

 

[NasuSama]

Sorry about this. Will anyone help? I showed some work.

 

[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?

 

[NasuSama]

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?

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)

 

[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. -__-

 

[NasuSama]

I actually got the solution by Newton's Law.

 

[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.

 

[NasuSama]

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.

That is actually the approach by energy considerations. Thanks!