Work/energy to find horizontal displacement

[jybe]

Homework Statement

A block slides down a frictionless incline (30 degrees above horizontal) for L=1.4m until it meets a horizontal surface with coefficient of kinetic friction 0.3 before coming to rest. Use work and energy to find the distance that the block slides on the horizontal surface before stopping.

L = 1.4 m
theta = 30 degrees
Coefficient of kinetic friction = 0.3
M = 3 kg

Homework Equations

mgh = 0.5mv²

The Attempt at a Solution

v = (2gh)^0.5
v = (2*9.8*1.4*sin(30))^0.5
v = 3.7 m/s (speed of block as it just enters the horizontal surface)

F(friction) = 0.3 * 2kg * 9.8
F(friction) = 5.88 N

a = F/m
a = 5.88 N /2 kg
a = 2.94 (opposite direction of velocity)

Vf² = Vi² + 2a*Δx
0 = 3.7² -(2*2.94*Δx)
Δx = 2.33 m

Therefore, the distance traveled on the horizontal surface is 2.33 m.

Does this look correct? I'm doubting the way I solved for acceleration.

 

[kuruman]

The idea is to use the work-energy theorem without going through the acceleration calculation. Can you do that? What does the work-energy theorem say?

 

[jybe]

The idea is to use the work-energy theorem without going through the acceleration calculation. Can you do that? What does the work-energy theorem say?

"According to the work-energy theorem, the net work on an object causes a change in the kinetic energy of the object. The formula for net work is net work = change in kinetic energy = final kinetic energy - initial kinetic energy."

But at some point I have to solve for displacement, so how can I do that without using kinematics?

 

[kuruman]

But at some point I have to solve for displacement, so how can I do that without using kinematics?

The whole point of the exercise is to show you exactly how to do that. Here's what you do, step by step.
1. Calculate the change in kinetic energy of the block and put it on the right side of the equation.
2. Figure out what and how many forces do work on the block as it is displaced.
3. Calculate the work done by each force on the block.
4. Add up all the works and put the sum on the left side of the equation.
5. Solve the equation for what you need to find.

On edit: I swapped "left" and "right" in the steps above to conform with your statement of the work-energy theorem.

 

[jybe]

The whole point of the exercise is to show you exactly how to do that. Here's what you do, step by step.
1. Calculate the change in kinetic energy of the block and put it on the right side of the equation.
2. Figure out what and how many forces do work on the block as it is displaced.
3. Calculate the work done by each force on the block.
4. Add up all the works and put the sum on the left side of the equation.
5. Solve the equation for what you need to find.

On edit: I swapped "left" and "right" in the steps above to conform with your statement of the work-energy theorem.

That makes sense, except I'm going to have F*d on one side ... and to solve for F, I'll need acceleration (F=ma) ? I can't seem to see my way through that.

 

[kuruman]

Have you completed step 2? How many forces do you have and what is the name of this F in your list of forces?

 

[jybe]

Have you completed step 2? How many forces do you have and what is the name of this F in your list of forces?

Change in kinetic energy = F(friction)*d
0.5mv² = (μ*mg)*d

There are no other forces acting on the block as far as I can tell.

 

[kuruman]

There are no other forces acting on the block as far as I can tell.

What about gravity and the incline?
Why is the change in kinetic energy 0.5 mv²? The change in kinetic energy is ΔK = E_final - E_initial. What is E_final? How about E_initial? The difference is a trivial number to calculate.

 

[jybe]

What about gravity and the incline?
Why is the change in kinetic energy 0.5 mv²? The change in kinetic energy is ΔK = E_final - E_initial. What is E_final? How about E_initial? The difference is a trivial number to calculate.

I can say that energy at the top of the ramp is equal to mgh, and when it gets to the bottom of the ramp, it will be 0.5mv², but still the same number because nothing has been lost to friction.

So now looking at the horizontal surface, KE final will be zero, and KE initial will be equal to mgh when the block was at the top of the ramp. KE is mgh, and on the other side of the equation, the only work being done is by friction

mgh = (μ*mg)*d
h = μ*d
d = h/μ

 

[kuruman]

... KE final will be zero ...

That's correct.

... KE is mgh ...

That's incorrect. The block starts from rest at the top of the ramp, so what is its initial KE?

... and on the other side of the equation, the only work being done is by friction

What about the work done by gravity and the work done by the ramp? Are they zero? Why?

 

[jybe]

That's incorrect. The block starts from rest at the top of the ramp, so what is its initial KE?

Well KE at the top of the ramp is zero, but KE initial (initial being the start of the horizontal plane) must be equal to mgh, because all the mgh was converted into KE...

What about the work done by gravity and the work done by the ramp? Are they zero? Why?

The ramp is frictionless, so there is no work done there, but the work done by gravity is mgh, right?

So just looking at the ramp, mgh = 0.5mv² (all potential energy converted to kinetic energy).

Looking at the horizontal plane now, all the block has is kinetic energy.

0.5mv² = F(friction)*displacement --> (initial kinetic energy is eventually converted into heat energy from friction)
0.5mv² = (μ*mg)*d

Isn't this right? Also, in my last post I had mgh equal to the work done by friction because mgh from the top of the ramp is equal to the kinetic energy at the bottom of the ramp, but I should have used KE to avoid confusion

 

[kuruman]

You are still missing the point of the exercise because you appear to be fixated on breaking the problem in two pieces, one that conserves mechanical energy from top to bottom of the ramp and the second piece that dissipates the kinetic energy while the potential energy does not change. The beauty of the work-energy theorem is that you don't have to do that. Consider the motion from point A = top of the ramp to point B = where the block stops. Ignore the transition point from no friction to friction. Answer the following questions
1. What is the change in kinetic energy from A to B?
2. What is the work done by gravity from A to B?
3. What is the work done by the incline from A to B?
4. What is the work done by friction from A to B?
Once you have answers to all four questions, add answers 2, 3 and 4 and set the sum equal to the answer in 1.

 

[jybe]

You are still missing the point of the exercise because you appear to be fixated on breaking the problem in two pieces, one that conserves mechanical energy from top to bottom of the ramp and the second piece that dissipates the kinetic energy while the potential energy does not change. The beauty of the work-energy theorem is that you don't have to do that. Consider the motion from point A = top of the ramp to point B = where the block stops. Ignore the transition point from no friction to friction. Answer the following questions
1. What is the change in kinetic energy from A to B?
2. What is the work done by gravity from A to B?
3. What is the work done by the incline from A to B?
4. What is the work done by friction from A to B?
Once you have answers to all four questions, add answers 2, 3 and 4 and set the sum equal to the answer in 1.

1) zero
2) mgh
3) zero
4) (μ*mg)*d

mgh + (μ*mg)*d = 0

 

[kuruman]

The work done by friction is negative because the angle between the displacement and the force is 180^o which makes the cosine -1. Otherwise it's correct. Now put the corrected equation together, solve for d, put in the numbers and you're done.

 

[jybe]

The work done by friction is negative because the angle between the displacement and the force is 180^o which makes the cosine -1. Otherwise it's correct. Now put the corrected equation together, solve for d, put in the numbers and you're done.

Thanks so much

mgh - (μ*mg)*d = 0
h/μ = d
d = (1.4*sin(30)) / 0.3
d = 2.3 m

 

[kuruman]

Good job. I hope you agree that this way is much quicker and less prone to mistakes than your original solution.

 

[jybe]

Good job. I hope you agree that this way is much quicker and less prone to mistakes than your original solution.

Absolutely!