# GCSE physics kinetic energy and work done question help needed!

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sammy_li11

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(https://qualifications.pearson.com/c...e_20130110.pdf - the question is in here somewhere)

there is a rollercoaster, launched from point A, accelerates to point B and then rises over point C. each loaded car has a mass of 2000kg and C is 128m above B

the car gains kinetic energy when work is done on it by the launching system between A and B. Assume there are no energy losses.

a) state the minimum kinetic energy that the car must have at B for it to reach C

the answer is 2.56mJ, which is the same as the gravitational potential energy gained by the car when it rises from B to C but i don't understand why it is the same

b) how is the kinetic energy gained related to the work done?

??? no clue on this one

could someone please explain these two questions? help would be appreciated!! thank youu

there is a rollercoaster, launched from point A, accelerates to point B and then rises over point C. each loaded car has a mass of 2000kg and C is 128m above B

the car gains kinetic energy when work is done on it by the launching system between A and B. Assume there are no energy losses.

a) state the minimum kinetic energy that the car must have at B for it to reach C

the answer is 2.56mJ, which is the same as the gravitational potential energy gained by the car when it rises from B to C but i don't understand why it is the same

b) how is the kinetic energy gained related to the work done?

??? no clue on this one

could someone please explain these two questions? help would be appreciated!! thank youu

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saranoo

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a) if im not wrong, i believe thats because in order for an object to keep at a constant velocity, it must have equal force acting backwards and forwards, if the backwards one were more, it would decelerate, (and if the forwards were more it would accelerate) hence the minimum for the car to keep going, is where forces going back and forwards are equal i.e. the minimum is where gravitional and kinetic forces are the same (as they act of different directions)

b) P= W/t OR P=E/t shows that work done=energy transfer therefore the kinetic energy transfer is proportional to work done

ngl i might be completely wrong, so soz if i am. good luck!

b) P= W/t OR P=E/t shows that work done=energy transfer therefore the kinetic energy transfer is proportional to work done

ngl i might be completely wrong, so soz if i am. good luck!

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Pangol

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(Original post by

(https://qualifications.pearson.com/c...e_20130110.pdf - the question is in here somewhere)

there is a rollercoaster, launched from point A, accelerates to point B and then rises over point C. each loaded car has a mass of 2000kg and C is 128m above B

the car gains kinetic energy when work is done on it by the launching system between A and B. Assume there are no energy losses.

a) state the minimum kinetic energy that the car must have at B for it to reach C

the answer is 2.56mJ, which is the same as the gravitational potential energy gained by the car when it rises from B to C but i don't understand why it is the same

b) how is the kinetic energy gained related to the work done?

??? no clue on this one

could someone please explain these two questions? help would be appreciated!! thank youu

**sammy_li11**)(https://qualifications.pearson.com/c...e_20130110.pdf - the question is in here somewhere)

there is a rollercoaster, launched from point A, accelerates to point B and then rises over point C. each loaded car has a mass of 2000kg and C is 128m above B

the car gains kinetic energy when work is done on it by the launching system between A and B. Assume there are no energy losses.

a) state the minimum kinetic energy that the car must have at B for it to reach C

the answer is 2.56mJ, which is the same as the gravitational potential energy gained by the car when it rises from B to C but i don't understand why it is the same

b) how is the kinetic energy gained related to the work done?

??? no clue on this one

could someone please explain these two questions? help would be appreciated!! thank youu

Not that you need it for this question, but it's useful to realise what will happen if the KE at B is less than, equal to or greater than 2.56 MJ. If it is less than this, the rollercoaster will not be able to get to C because it would need to have more GPE at C than the KE it starts off with. If it is equal, then the rollercoaster could get to C, but then all of the KE would have turned into GPE and it would have to stop at C. If it is more, then it could get to C and have some KE left over - in other words, it would still be moving at this point.

For (b), you need to revise what the connection between work done and energy transferred is. They are the same thing - if we do work on an object, we transfer energy to it. For example, if I throw a ball, I do work on it while I am throwing it (I apply a force to it that moves through the distance it moves while I am throwing it), and when it leaves my hand, it now has kinetic energy. This energy is equal to the amount of work I did on it.

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sammy_li11

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(Original post by

Part (a) is a conservation of energy question. The principle of conservation of energy says that energy cannot be destroyed or created, but can be changed from one form to another. You are told to assume that there are no energy losses, which means that the only way that energy changes once the rollercoaster has got going is between kinetic energy and gravitational potential energy. If you have worked out the GPE at C to be 2.56 MJ (I assume you mean MJ rather than mJ), then the KE at B will have to be at least this - otherwise, there will be some GPE at C which has come from nowhere.

Not that you need it for this question, but it's useful to realise what will happen if the KE at B is less than, equal to or greater than 2.56 MJ. If it is less than this, the rollercoaster will not be able to get to C because it would need to have more GPE at C than the KE it starts off with. If it is equal, then the rollercoaster could get to C, but then all of the KE would have turned into GPE and it would have to stop at C. If it is more, then it could get to C and have some KE left over - in other words, it would still be moving at this point.

For (b), you need to revise what the connection between work done and energy transferred is. They are the same thing - if we do work on an object, we transfer energy to it. For example, if I throw a ball, I do work on it while I am throwing it (I apply a force to it that moves through the distance it moves while I am throwing it), and when it leaves my hand, it now has kinetic energy. This energy is equal to the amount of work I did on it.

**Pangol**)Part (a) is a conservation of energy question. The principle of conservation of energy says that energy cannot be destroyed or created, but can be changed from one form to another. You are told to assume that there are no energy losses, which means that the only way that energy changes once the rollercoaster has got going is between kinetic energy and gravitational potential energy. If you have worked out the GPE at C to be 2.56 MJ (I assume you mean MJ rather than mJ), then the KE at B will have to be at least this - otherwise, there will be some GPE at C which has come from nowhere.

Not that you need it for this question, but it's useful to realise what will happen if the KE at B is less than, equal to or greater than 2.56 MJ. If it is less than this, the rollercoaster will not be able to get to C because it would need to have more GPE at C than the KE it starts off with. If it is equal, then the rollercoaster could get to C, but then all of the KE would have turned into GPE and it would have to stop at C. If it is more, then it could get to C and have some KE left over - in other words, it would still be moving at this point.

For (b), you need to revise what the connection between work done and energy transferred is. They are the same thing - if we do work on an object, we transfer energy to it. For example, if I throw a ball, I do work on it while I am throwing it (I apply a force to it that moves through the distance it moves while I am throwing it), and when it leaves my hand, it now has kinetic energy. This energy is equal to the amount of work I did on it.

(Original post by

a) if im not wrong, i believe thats because in order for an object to keep at a constant velocity, it must have equal force acting backwards and forwards, if the backwards one were more, it would decelerate, (and if the forwards were more it would accelerate) hence the minimum for the car to keep going, is where forces going back and forwards are equal i.e. the minimum is where gravitional and kinetic forces are the same (as they act of different directions)

b) P= W/t OR P=E/t shows that work done=energy transfer therefore the kinetic energy transfer is proportional to work done

ngl i might be completely wrong, so soz if i am. good luck!

**saranoo**)a) if im not wrong, i believe thats because in order for an object to keep at a constant velocity, it must have equal force acting backwards and forwards, if the backwards one were more, it would decelerate, (and if the forwards were more it would accelerate) hence the minimum for the car to keep going, is where forces going back and forwards are equal i.e. the minimum is where gravitional and kinetic forces are the same (as they act of different directions)

b) P= W/t OR P=E/t shows that work done=energy transfer therefore the kinetic energy transfer is proportional to work done

ngl i might be completely wrong, so soz if i am. good luck!

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randomusername23

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Yes - you work out GPE using Ep = mgh = 2000kg * 9.81m/s^2 * 128m etc.

Part b) is basically about how work done on the object in the form of a force gives the object kinetic energy etc. And yes, Power = Energy Transferred / TIme is the same thing as P = work done / time. They are equivalent - you are transferring energy to the object.

Part b) is basically about how work done on the object in the form of a force gives the object kinetic energy etc. And yes, Power = Energy Transferred / TIme is the same thing as P = work done / time. They are equivalent - you are transferring energy to the object.

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Pangol

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I would say that looking at part (b) in terms of equations is uneccesary and potentially confusing. It is about the ideas and definitions of the subject. Work done and energy transferred are the same thing - that's all you need.

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randomusername23

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Pangol

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(Original post by

The equations help understanding, but yes it's more theoretical than mathematical

**AndrewMarriott3**)The equations help understanding, but yes it's more theoretical than mathematical

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randomusername23

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It only demonstrates that work done and energy transferred are the same thing. It was just the first thing that came to mind when connecting the 2

(Original post by

I don't think that the equations quoted help to clarify anything, though. There is nothing in the question that involves using power or time. It is all about work done and energy transferred. The equations are not wrong, but why bring power into it?

**Pangol**)I don't think that the equations quoted help to clarify anything, though. There is nothing in the question that involves using power or time. It is all about work done and energy transferred. The equations are not wrong, but why bring power into it?

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(Original post by

It only demonstrates that work done and energy transferred are the same thing. It was just the first thing that came to mind when connecting the 2

**AndrewMarriott3**)It only demonstrates that work done and energy transferred are the same thing. It was just the first thing that came to mind when connecting the 2

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Yes, I see where you are coming from. TBH though, if you know and understand the equations you can get half the marks on a paper without knowing any of the theory at all. Not that it's relevant or that anyone would do that though.

(Original post by

Well fair enough, and you are of course not wrong. Please don't think I'm trying to have a go at you! My worry is that a lot of people have problems with physics because they think it's just a load of equations and they find this confusing. In this case, part (b) doesn't need any equations - it's just to do with the idea that work done and energy transferred are the same thing.

**Pangol**)Well fair enough, and you are of course not wrong. Please don't think I'm trying to have a go at you! My worry is that a lot of people have problems with physics because they think it's just a load of equations and they find this confusing. In this case, part (b) doesn't need any equations - it's just to do with the idea that work done and energy transferred are the same thing.

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