Work/energy to find horizontal displacement

• jybe
In summary: 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
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

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.

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?

kuruman said:
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?

jybe said:
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.

kuruman said:
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.

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

kuruman said:
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.

jybe said:
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 = Efinal - Einitial. What is Efinal? How about Einitial? The difference is a trivial number to calculate.

kuruman said:
What about gravity and the incline?
Why is the change in kinetic energy 0.5 mv2? The change in kinetic energy is ΔK = Efinal - Einitial. What is Efinal? How about Einitial? 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

jybe said:
... KE final will be zero ...
That's correct.
jybe said:
... KE is mgh ...
That's incorrect. The block starts from rest at the top of the ramp, so what is its initial KE?
jybe said:
... 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?

kuruman said:
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...

kuruman said:
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

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.

kuruman said:
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

The work done by friction is negative because the angle between the displacement and the force is 180o 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.

kuruman said:
The work done by friction is negative because the angle between the displacement and the force is 180o 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

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

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

What is work?

Work is defined as the product of force and displacement. In other words, it is the amount of energy transferred when a force is applied to an object and causes it to move a certain distance in the direction of the force.

What is energy?

Energy is the ability to do work. It comes in many forms, such as kinetic energy (energy of motion) and potential energy (stored energy). In the context of work, energy is transferred from one object to another when work is done.

How is work related to energy?

Work and energy are closely related concepts. Work is the transfer of energy from one object to another, while energy is the ability to do work. In other words, work is a measure of energy transfer.

What is the equation for calculating work?

The equation for calculating work is W = F * d, where W is work, F is the force applied, and d is the displacement in the direction of the force. This equation only applies to situations where the force and displacement are in the same direction.

How can work/energy be used to find horizontal displacement?

In order to find horizontal displacement using work and energy, you need to know the initial and final energies of an object, as well as the work done on the object. By using the equation W = ΔKE + ΔPE, where W is work, ΔKE is the change in kinetic energy, and ΔPE is the change in potential energy, you can solve for the displacement of the object in the horizontal direction.

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