Lower Secondary Science

[DJ123]

Thank you

[DJ123]

Thank you

[Verysuperkiasu]

Hi Dr Daniels…


Since the unit for energy is in Joules (for eg potential energy of a ball is 50 Joules) then is this energy also equals to the workdone to the ball? In which case you do not need to apply the formula Workdone=Force x Dist moved in the same direction?

[Dr.033430Daniel]

Verysuperkiasu:

Hi Dr Daniels..


Since the unit for energy is in Joules (for eg potential energy of a ball is 50 Joules) then is this energy also equals to the workdone to the ball? In which case you do not need to apply the formula Workdone=Force x Dist moved in the same direction?

I’ll answer with some specific numbers.

If a ball with a mass of 1kg has a gravitational potential energy of 50J, then using the formula for potential energy,
mgh = 50 J
(1kg) (10m/s2) h = 50 J

Solving for h the height is 5 meters. So this ball is 5 meters high.
Suppose a question asks how much work is needed to raise the ball to a level of 5m (assuming no air friction). The answer would be 50 Joules.

Putting this in a formula, the force needed to move the ball the distance of 5m would be a force equal to the weight of the ball. Force is mg.

F = mg = (1kg) (10m/s2) = 10N

W = 10N (5m) = 50J

A force of 10N acting over a distance of 5m would raise the ball at constant velocity and now the ball would have 50J of gravitational potential energy.
Thinking in terms of the law of conservation of energy, the work done on the ball goes into an increase of gravitational potential energy. If I used a greater force and input more work, then the ball would gain kinetic energy in addition to the gravitational potential energy.

[BoboPanda]

Hi, hi, wonder whether anyone of you can help me? I have been following this topic "Secondary School Science" under the Secondary Schools - Academic Support for quite a while even though I have not registered with this forum. I found the questions posed and answers provided very helpful. Lately, I can’t find this topic under this "Secondary Schools - Academic Support". Can anyone tell me whether the topic is still around? If so, where it is? Very upset that I can’t find it anymore.

[jedamum]

Went to 'happenings' section.

Maybe admin can rectify
http://www.kiasuparents.com/kiasu/forum ... =43&t=7748

[BoboPanda]

Thank you, Jedamum, for your prompt reply and the link to the "happenings" section.


I will try to contact admin to move it back to the Secondary Schools - Academic Support. Hope the admin will move it back.

[Verysuperkiasu]

Dr.Daniel:

Putting this in a formula, the force needed to move the ball the distance of 5m would be a force equal to the weight of the ball. Force is mg.

F = mg = (1kg) (10m/s2) = 10N

W = 10N (5m) = 50J

But I thought the force needed to move the ball has to be an external force applied to the ball which is greater than the weight of the ball to move it upwards? coz if the force applied is equal to the weight of the ball, they cancel each other out and the ball can't move?

Secondly, in this case, the force is applied upwards. If the ball is kicked horizontally, Workdone = force X dist moved in same direction, then we can't take weight of the ball into account can we? since weight is downwards due to gravity and is not the same direction as the horizontal motion?

Dr.Daniel:

A force of 10N acting over a distance of 5m would raise the ball at constant velocity and now the ball would have 50J of gravitational potential energy.
Thinking in terms of the law of conservation of energy, the work done on the ball goes into an increase of gravitational potential energy. If I used a greater force and input more work, then the ball would gain kinetic energy in addition to the gravitational potential energy.

In other words, can we say that the change in potential energy is the same as the work done on the object?

Actually I came across this question (below):

A rock lifted up to the top of the cliff 120m from the ground has a potential energy of 12kJ. How much work has been done to lift the rock?
a) 0J b)10kJ c)12kJ d)1440kJ

Can't use formula workdone = force x dist moved as we don't know the force applied. And since the answer is c)12kJ, I thought maybe P.E. is the same as work done (afterall they're both in Joules)...that's what led me to raise this question in the first place.

[sunkc2000]

My dd just gotten back her Sec 2 term 1 test paper. Realised she did badly for her open ended questions. I read through her papers and found that she, probably has the knowledge, but phrasing the answers were a challenge to her. Hence, she was not able to score full marks for most for the questions.


Can anyone advise how can she improve in that area?
Does teacher look at Key words (like Pri school) when marking the papers?

[Dr.033430Daniel]

Verysuperkiasu:

But I thought the force needed to move the ball has to be an external force applied to the ball which is greater than the weight of the ball to move it upwards? coz if the force applied is equal to the weight of the ball, they cancel each other out and the ball can't move?

If an upwards force is greater than the weight of the ball, then the ball will accelerate upwards. If the upwards force is equal to the weight of the ball then it could be true that 1. the ball is not moving or 2. that the ball is moving at a constant velocity.

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Sorry it took me a while to find this thread again - why was this thread moved out of the academic support area?