Energy and bringing an ibject up an incline

[sgstudent]

If I have a 2 box with the same weight and I am to bring it up two inclines with different heights but different level of steepness. Will the work done to bring the object up be the same? Because if the incline have the same surface then it has the same amount of friction. Logical it would seem that the less steep incline will have the greater work done to bring in up as the amounts if work done against friction will be greater. But then again, it would also seem that to push something up from ground to the higher level from a steep incline then work done is required. But I'm not sure about this. Thanks for the help!

 

[Opus_723]

The amount of work done will not depend on the steepness of the incline, but it will depend on the height. But you are right, it will seem easier to move the box on the less steep ramp.

The explanation is that although you exert less force to move the box on the shallow ramp, you have to move it a greater distance.

I suggest drawing some diagrams and calculating the forces required and the work done in moving a box up a ramp. You will find that the force you exert to move the box depends on the angle, but the work done on the box does not.

 

[sgstudent]

The amount of work done will not depend on the steepness of the incline, but it will depend on the height. But you are right, it will seem easier to move the box on the less steep ramp.

The explanation is that although you exert less force to move the box on the shallow ramp, you have to move it a greater distance.

I suggest drawing some diagrams and calculating the forces required and the work done in moving a box up a ramp. You will find that the force you exert to move the box depends on the angle, but the work done on the box does not.

But is there a energy explanation for this? I understand the concept using the vector diagram but then again I don't understand using just energy. Or is the energy based on the force diagram. It's because of the angle that the friction acts and also the normal force's changed direction resulting in a greater force? Also I'm unsure about questions where they ask to 'stop a moving object' if the object has a certain amount if kinetic energy at that point, then to stop it I need an work done by brakes which is the same. But if I simply have a force to stop it will it stop?

And if I push a box with 10N over a distance of 5m will the energy at the end be 50N? This includes resisting forces such as friction.

Thanks for the help!

 

[emailanmol]

Just to clarify if a body is moved through a certain height then work is being done:
1.Against any resistive forces such as friction.
2.To increase any kinetic energy the body may acquire.
3.To increase the gravitational potential energy.

It is the increase of potential energy which is independent of the method used to lift the body and the route it takes.

...and therefore a method (like a lower slope) that increases friction also increases the required input energy to attain the same height.

Hey, just to clarify
Is the equation i wrote correct ?
That suggest work actually depends on factor (uR+H)mg

Where R is base length and Height is H.

Infact this is valid even for a mountain like path(which infact consists of small inclines of different angles of inclination)

Isn't it?

What I don't get is friction is coming independent of path.
It is coming same for two different mountains which have same R and H but different L.


(provided we keep belocity 0 all through out the path)

Here is that post.

Hey, I maybe wrong but IMHO the answer depends on both height and on the base length of the incline.

I can throw in some equations to back up my reasoning .

At every stage (assuming 0 acceleration as we are moving the blocks slowly.)

F=mgsin(c) +umgcos(c)

where c is angle of inclination.

Work done by all forces is 0 as no change in KE takes place.

Suppose the baselength is R, height is h and hypotenuse us L.

So work done by F
w=integration (mgsin(c)dl + ungcos(c) dl)

What is dlsin(c) and dlcos(c)?

What does the integration fetch.

 

[emailanmol]

But is there a energy explanation for this? I understand the concept using the vector diagram but then again I don't understand using just energy. Or is the energy based on the force diagram. It's because of the angle that the friction acts and also the normal force's changed direction resulting in a greater force? Also I'm unsure about questions where they ask to 'stop a moving object' if the object has a certain amount if kinetic energy at that point, then to stop it I need an work done by brakes which is the same. But if I simply have a force to stop it will it stop?

And if I push a box with 10N over a distance of 5m will the energy at the end be 50N? This includes resisting forces such as friction.

Thanks for the help!

Hey,

You seem to be having some problem in understanding and relating work (and thus energy )and force.

And if I push a box with 10N over a distance of 5m will the energy at the end be 50N? This includes resisting forces such as friction.

First of all energy is frame dependent.

Saying something posseses 50J of energy is not right.

However, if a 2N force moves a body by 3m(in absence of all other forces) we say the body now posses an extra energy of 6J.

Coming to your question if you push the block with 10N force for 5m , work done by that force is 50J.

This may or maybot be the energy transferred to the block .

Suppose another force acts opposite to direction of motion with value 3N.

Work done by this force on block is -15J.

So the blocks energy increased by (50-15) 35 joules.

Energy received by block is the net change in its energy i.e the result of net work done by all EXTERNAL forces acting on it.

(As you will study higher, this statement (referring to only EXTERNAL forces)is strictly valid for only a particle.For a system of particles energy can also be changed by change on internal composition which may take place due to internal forces)

 

[sgstudent]

Hey,

You seem to be having some problem in understanding and relating work (and thus energy )and force.




First of all energy is frame dependent.

Saying something posseses 50J of energy is not right.

However, if a 2N force moves a body by 3m(in absence of all other forces) we say the body now posses an extra energy of 6J.

Coming to your question if you push the block with 10N force for 5m , work done by that force is 50J.

This may or maybot be the energy transferred to the block .

Suppose another force acts opposite to direction of motion with value 3N.

Work done by this force on block is -15J.

So the blocks energy increased by (50-15) 35 joules.

Energy received by block is the net change in its energy i.e the result of net work done by all EXTERNAL forces acting on it.

(As you will study higher, this statement (referring to only EXTERNAL forces)is strictly valid for only a particle.For a system of particles energy can also be changed by change on internal composition which may take place due to internal forces)

Then is the formula net work done=KE final-KE initial correct? If I apply a force over a distance then work is being done. Then how much energy will the thing have? Using the formula if my original velocity is 0 then the final kinetic energy is the same work done. But in a case where there is an incline but I don't know the height of the inclination then just by using work done can I get my final energy? Or should I be able to get the GPE along the way or else it will be too complicated to solve?

Also when I have a roller coaster with a kinetic energy of 100J, then to stop it I need 100J of braking energy. But won't it have just 0 net work done. So by net work done=KE final-KE initial. Then won't the final KE have to be the same as the initial KE? Thanks for the help!

 

[emailanmol]

Also I'm unsure about questions where they ask to 'stop a moving object' if the object has a certain amount if kinetic energy at that point, then to stop it I need an work done by brakes which is the same. But if I simply have a force to stop it will it stop?

It depends on if the force does work and how it does work.

To picture my example clearly, imagine a small plane which going both up and towards right (lets say at an angle 45 degrees with horizontal)(imagine gravity free space )

Let its mass be 2kg and its velocity in y direction be 2m/s and in x be 2m/s (its moving at 45 degrees).

Its kinetic energy total is 8J ,

(4J due to movement in y and 4J due to movement in x.Although we never assosciate energy with direction, i am stating this ti make my example clear)

Now suppose you apply a force with your hand in negative y direction
The max (negative) amount of work this force can do is -4J.

By doing work of -4J this force brings the plane to halt in y direction.
Any further work done by this force will result in positive work now because the displacement in -y direction will be parallel to this force.

To stop the body completely you will have to apply a force along x direction as well and do a work of -4 J.

Also you were talking of a car traveling and you applying a force to stop it.
At this point the car has positive velocity.
Its velocity can be brought to 0 only when you decelerate it in oppsing direction with a force.(which is what we saw by the energy explanation that a force has to act opposite to velocity vector(remember velocity vector is always parallel ti displacement vector) to stop it.any force in perpendicular direction will be of no use as this won't cause a deceleration nor will it be capable of taking away all kinetic energy of the body.


In essence the kinematics and energy descriptions are one and the same and are just two different ways to see a physical situation.

Using one over the other is simply a matter of convenience.

 

[emailanmol]

You are making a simple mistake of using energy possesed by body in equations.


Work energy theorem is correct and it states that net work done on body is change in its kinetic energy

Work done by braking force (call it w)on roller coaster is what we will find (and it will come out as i-100J and not 0. Work done is energy transferred.to stop the coaster work has to be done )

Its intial kinetic energy is -100J.

Let its final kinetic energy be x.(we know it should be 0 for coaster to stop)

From work energy theorem
W=Ke(final)-Ke(initial)


W=x-(100)
W=0-100

So w is -100 J.

I.e work done is -100 J to stop it

 

[emailanmol]

Then is the formula net work done=KE final-KE initial correct? If I apply a force over a distance then work is being done. Then how much energy will the thing have? Using the formula if my original velocity is 0 then the final kinetic energy is the same work done. But in a case where there is an incline but I don't know the height of the inclination then just by using work done can I get my final energy? Or should I be able to get the GPE along the way or else it will be too complicated to solve!

You are absolutely right.Unless you know about kinetic energy at top applying work energy theorem will give you nothing.

Therefore

The problem should be stated like this

"A block has to carried on two different inclines having base lengths
R1, R2 and heights H1,H2.

The block is carried slowly to the top.
In which case work done by external force will be more."


The term taken slowly has to be there else the question can't be solved.

It implies kinetic energy is nearly 0 all throughout the motion.(and therefore at top most point)

Otherwise you cannot find the energy at top.

(see my first post and tell if you understand the solution)



I hope I am making it ckear enough for you :-)

 

[Dadface]

Just to clarify if a body is moved through a certain height then work is being done:
1.Against any resistive forces such as friction.
2.To increase any kinetic energy the body may acquire.
3.To increase the gravitational potential energy.

It is the increase of potential energy which is independent of the method used to lift the body and the route it takes.

 

[russ_watters]

...and therefore a method (like a lower slope) that increases friction also increases the required input energy to attain the same height.

 

[emailanmol]

Just to clarify if a body is moved through a certain height then work is being done:
1.Against any resistive forces such as friction.
2.To increase any kinetic energy the body may acquire.
3.To increase the gravitational potential energy.

It is the increase of potential energy which is independent of the method used to lift the body and the route it takes.

...and therefore a method (like a lower slope) that increases friction also increases the required input energy to attain the same height.

Hey, just to clarify
Is the equation i wrote correct ?
That suggest work actually depends on factor (uR+H)mg

Where R is base length and Height is H.

Infact this is valid even for a mountain like path(which infact consists of small inclines of different angles of inclination)

Isn't it?

What I don't get is friction is coming independent of path.
It is coming same for two different mountains which have same R and H but different L.


(provided we keep belocity 0 all through out the path)

Here is that post.

Hey, I maybe wrong but IMHO the answer depends on both height and on the base length of the incline.

I can throw in some equations to back up my reasoning .

At every stage (assuming 0 acceleration as we are moving the blocks slowly.)

F=mgsin(c) +umgcos(c)

where c is angle of inclination.

Work done by all forces is 0 as no change in KE takes place.

Suppose the baselength is R, height is h and hypotenuse us L.

So work done by F
w=integration (mgsin(c)dl + ungcos(c) dl)

What is dlsin(c) and dlcos(c)?

What does the integration fetch.

 

[Opus_723]

With being the angle of the incline and h being the height of the incline...

Force required to push the box: F = mgsinθ + μmgcosθ (you got this right).

Work done on box: F*distance = F*(h/sinθ) = (mgsinθ)*(h/sinθ) + (μmgcosθ)*(h/sinθ)

= mgh + μmgh/tanθ

The work is converted to potential energy (through the height change) and thermal energy (lost to friction), not kinetic energy, so you cannot use the change in kinetic energy to find the work done.

 

[sgstudent]

With being the angle of the incline and h being the height of the incline...

Force required to push the box: F = mgsinθ + μmgcosθ (you got this right).

Work done on box: F*distance = F*(h/sinθ) = (mgsinθ)*(h/sinθ) + (μmgcosθ)*(h/sinθ)

= mgh + μmgh/tanθ

The work is converted to potential energy (through the height change) and thermal energy (lost to friction), not kinetic energy, so you cannot use the change in kinetic energy to find the work done.

I saw this on physicsclassroom. There's a formula for it Work done by applied force=force