Physics - Work done
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A person exerts a horizontal force of 500 N on a box, which also experiences a friction force of 100 N. How much work is done against friction when the box moves a horizontal direction of 3 m?
Do you get the resultant force and multiply by 3 m? 400 X 3 = 1200? I don't really get the question when it says "work done against friction".
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Work done against friction = energy lose due to friction
So work done against friction = 100 x 3 = 300J
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Originally posted by Uncertain:
Work done against friction = energy lose due to friction
So work done against friction = 100 x 3 = 300J
thanks so much. i was thinking about that too! by the way, can you tell me why a hydroelectric generator device wastes energy? the question goes something like "the water supplies 2000J of energy every second to the wheel. the electrical energy output is 1200 J every second.1 way which it wastes energy is perhaps through the loss of thermal energy i suppose. during the conversions of different forms of energy, there is bound to be some energy which is lost as heat energy. what is the second reason? i can't think of another.
and there;s this question. if is of a car moving on a horizontal section until it reaches a ramp and moves downwards. length of the ramp is 0.9 m .
(ii) the car has 0.30 J of kinetic energy at the topof the ramp and loses 0.50 of potential energy as it moves to the bottom of the ramp. calculate kinetic energy of car at bottom of the ramp.
i tried to equate using principle of conservation of energy.
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wad i think is
1) energy required to overcome the inertia of the wheel as the wheel is likely to be large.
2)friction between water molecules and the wheel.
2nd question... 0.3 x 0.5= 0.15J= KE at bottom (ans)
on top, PE = KE = 0.3
as u go down to bottom lose 1/2 of it.
at bottom, energy converted to KE again.
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Originally posted by arigatoast:
wad i think is
1) energy required to overcome the inertia of the wheel as the wheel is likely to be large.
2)friction between water molecules and the wheel.
2nd question... 0.3 x 0.5= 0.15J= KE at bottom (ans)
on top, PE = KE = 0.3
as u go down to bottom lose 1/2 of it.
at bottom, energy converted to KE again.
Eh, why is it that on the top PE = KE? I don't think I've ever heard of that before. Can anyone explain?
I only know a scenario when there is G.P.E on top but no K.E and vice versa for the bottom.
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Originally posted by bonkysleuth:
and there;s this question. if is of a car moving on a horizontal section until it reaches a ramp and moves downwards. length of the ramp is 0.9 m .
(ii) the car has 0.30 J of kinetic energy at the topof the ramp and loses 0.50 of potential energy as it moves to the bottom of the ramp. calculate kinetic energy of car at bottom of the ramp.
i tried to equate using principle of conservation of energy.
At top of ramp, car has KE (0.30J) and GPE (this is equal to 50J since the car loses 50J of GPE when it travels to the bottom).
Assuming no energy is lost as thermal energy, all GPE is converted to KE.
Thus KE at bottom of ramp = 80J?
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Originally posted by bonkysleuth:
thanks so much. i was thinking about that too! by the way, can you tell me why a hydroelectric generator device wastes energy? the question goes something like "the water supplies 2000J of energy every second to the wheel. the electrical energy output is 1200 J every second.1 way which it wastes energy is perhaps through the loss of thermal energy i suppose. during the conversions of different forms of energy, there is bound to be some energy which is lost as heat energy. what is the second reason? i can't think of another.
and there;s this question. if is of a car moving on a horizontal section until it reaches a ramp and moves downwards. length of the ramp is 0.9 m .
(ii) the car has 0.30 J of kinetic energy at the topof the ramp and loses 0.50 of potential energy as it moves to the bottom of the ramp. calculate kinetic energy of car at bottom of the ramp.
i tried to equate using principle of conservation of energy.
Ok I think eagle will kill me if i keep on taking his lobang... hehe BUT he is the best in physics here...Nevertheless, these are my thoughts...
(i) 1) Energy is needed to overcome the inertia of the wheel so energy is wasted.
2) Like what u said.... energy is lost in form of sound, heat and so on energy.
ii) is 0.50J of PE or 1/2 of PE loss? Be accurate in ur question.
if 0.50J, then follow secretliker
if 1/2, then follow agriatoast.
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Originally posted by Uncertain:
Ok I think eagle will kill me if i keep on taking his lobang... hehe BUT he is the best in physics here...Nevertheless, these are my thoughts...
(i) 1) Energy is needed to overcome the inertia of the wheel so energy is wasted.
2) Like what u said.... energy is lost in form of sound, heat and so on energy.
ii) is 0.50J of PE or 1/2 of PE loss? Be accurate in ur question.
if 0.50J, then follow secretliker
if 1/2, then follow agriatoast.
Dun worry, just take
I quite busy, and a bit sick...
In homework forum, as long as you can do the question and explain, just go ahead! We are sharing afterall, no difference
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Originally posted by secretliker:
At top of ramp, car has KE (0.30J) and GPE (this is equal to 50J since the car loses 50J of GPE when it travels to the bottom).
Assuming no energy is lost as thermal energy, all GPE is converted to KE.
Thus KE at bottom of ramp = 80J?
Hi secret liker.. how come is 80J? the energy mentioned are KE 0.3J and 0.5 of GPE lost. U mean 0.8 J?
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Originally posted by bonkysleuth:
Eh, why is it that on the top PE = KE? I don't think I've ever heard of that before. Can anyone explain?
I only know a scenario when there is G.P.E on top but no K.E and vice versa for the bottom.
oops sorry.. write wrongly.
on top of the ramp.. just b4 it slide down Max PE with 0.3J KE as stated in question.
i think u mean lose 0.5 J of GPE instead of 0.5 of GPE. so lost in GPE = gain in KE=0.5J
0.5J+0.3J = 0.8J (Ans)
ignore my previous post. thanks.
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Yup. It's 0.8J (I carelessly made it 80J).