# Solving Work Done Incline Homework Equations

• Sarah00
In summary, the conversation discusses two problems involving the calculation of work and the use of the principles of energy. In the first problem, the force is equal to the opposing gravitational force, and the work done is equal to the force multiplied by the distance traveled up the incline. In the second problem, there is a change in kinetic energy and potential energy, and the work done can be calculated by considering these changes. However, this is a more complex problem than the first one.
Sarah00

W = F d
F = ma

## The Attempt at a Solution

Constant Speed ==> a = 0 ==> Net Force = 0 ==> F cos 37 = mg sin 37 ==> F = 37.6 N
sin 37 = h/d ==> 6/10 = 0.5/d ==> d = 5/6
W = F d = 5/6 * 37.6 = 31.3 N [Not In Choices] - Any Help?

Perhaps it would be better to consider the work done as the change in potential energy and approach it that way.

yup! W = mgh = 5 * 10 * 0.5 = 25 J
In my previous attempt, my mistake was not writing cos 37 in finding the work

Sir,
I applied the method you mentioned on this questions but it does not work with me!

Confused now on when to use that method

That method find work done by gravity, is it same as work done by F ? Why?

Sarah00 said:
Sir,
I applied the method you mentioned on this questions but it does not work with me!

Confused now on when to use that method

That method find work done by gravity, is it same as work done by F ? Why?
The original problem had constant speed. I think you will find that this one doesn't.

Your initial method (after correcting by including cos(37°) in that case) should work here.

How "constant speed" differ?

Sarah00 said:
How "constant speed" differ?

KE is constant.

Sorry, but how is that relevant?
The way I think of the work done is final energy - initial energy
In final energy I have potential energy (mgh) only
In initial energy there is nothing in this case
So,
W = mgh

Where is my point of confusion?

The final energy will not be only in the form of potential energy.

Sarah00 said:
That method find work done by gravity, is it same as work done by F ? Why?
Without thinking too much, why don't you attempt this question using your previous knowledge on basics...

ok, let say the speed is not constant..
E initial is zero because potential is zero and it is at rest at the begining
E final is potential and kinetic energy will be ..

That also the same with if speed is constant.

So, why it differ

I did attempt the question and got the answer of 52 J but I am asking why I can't use that method

If the speed is constant to a direction, it means it has no external unbalancrd force towards that direction..

When there's an unbalanced force and the object shows a displacement, there's a work done.

That's right, but it does not answer my question in that why I can't use the concept of conservation of energy

We don't know the final kinetic energy. Since there is an external unbalanced force there is an acceleration so the initial KE cannot be taken as the final KE. So to apply conservation of energy, we don't have enough data.

OK Thanks, what about if it is constant speed. Initially velocity = 0 but final is not 0. So how we only used W = change in potential energy, without considering hte KE

Sarah00 said:
OK Thanks, what about if it is constant speed. Initially velocity = 0 but final is not 0. So how we only used W = change in potential energy, without considering hte KE
If there is a motion I think there should be a velocity.;)

Sarah00 said:
Sorry, but how is that relevant?
The way I think of the work done is final energy - initial energy
In final energy I have potential energy (mgh) only...

That is correct for #1 (constant velocity) but not #4 (constant force)

In #4 the block is accelerating so the final KE is not the same as the initial KE.

You can use conservation of energy to solve this but you would need to calculate the final velocity.

It is easier to use work = force * displacement.

Thanks, but why you said it is correct for #1 ?
It is constant velocity so we still have velocity in Ef but we don't have it in Ei (before applying the force)

Guys, I understand that in [constant force] we can't use principle of energy without find final KE.
But my problem is why we could use it in [constant speed], as we still have final KE

Sarah00 said:
Guys, I understand that in [constant force] we can't use principle of energy without find final KE.
But my problem is why we could use it in [constant speed], as we still have final KE
W = ΔE :
(Work done by an external force) = Change in total mechanical energy.
also E = KE + PE
So W = ΔKE + ΔPE​

In the first problem:
The velocity was constant, so the Kinetic Energy didn't change. Therefore, the work done by force, F, was equal to the change in Potential Energy.

When you did the problem using the definition of Work, you had to find the amount of force which produced zero acceleration. -- It was just enough to counteract gravity.​

In the second problem:
You are given the force. It's unlikely that it's the exact amount to counteract gravity, so the acceleration will not be zero. There will be a change in kinetic energy as well as in potential energy. You can still get an answer from energy considerations, but it will be quite a bit more involved than simply calculating the Work directly.​

But isn't the speed constant DURING the action of the force NOT before? So before (E initial) there was no movement i.e. KEi = 0 ?

Sarah00 said:
But isn't the speed constant DURING the action of the force NOT before? So before (E initial) there was no movement i.e. KEi = 0 ?
That issue is not addressed in the problem. (You're referring to the first problem, right?)

It simply says that the block is moving up the incline at a constant speed.

Aha now I am happy enough..

Same kinetic energy before and after so they cancel each other

Thanks guys !
You were so kind !

Problem 1
the driving force (f) = the opposing gravitational force
the work done = f * distance traveled up the incline
Problem 2
The change in KE, the gain in PE and the work done all have the same value

dean barry said:
...

Problem 2
The change in KE, the gain in PE and the work done all have the same value
That's not correct for problem 2.

Your right, the work done = ( gain in KE + gain in PE ), according to my calcs the work done = 111.81 Joules

## What is the formula for calculating work done on an incline?

The formula for calculating work done on an incline is W = Fd cosθ, where W is the work done, F is the applied force, d is the distance over which the force is applied, and θ is the angle between the force and the direction of motion.

## How do you calculate the force required to lift an object up an incline?

To calculate the force required to lift an object up an incline, you can use the formula F = mg sinθ, where F is the required force, m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the incline.

## What is the difference between work done and power?

Work done is the amount of energy transferred by a force acting over a certain distance, while power is the rate at which work is done or the amount of work done per unit of time. In other words, work done tells us the total amount of energy used, while power tells us how quickly that energy is being used.

## How does the angle of inclination affect the amount of work done?

The angle of inclination affects the amount of work done because it changes the distance over which the force is applied. The steeper the incline, the shorter the distance, and therefore less work is done. On the other hand, a smaller angle of inclination means a longer distance and more work done.

## Can you determine the work done on an incline without knowing the angle?

Yes, you can determine the work done on an incline without knowing the angle if you have information about the force applied and the distance over which it is applied. You can use the formula W = Fd cosθ and solve for θ using basic algebra. Alternatively, you can use the formula W = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the incline.

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