I know work is a scalar quantity, but this question came into my mind and I was not entirely sure. I'm guessing yes, because I applied a force over a distance, but I don't think I would have changed the kinetic energy of the object if I bring it to a stop at the end. Any answers would be appreciated.
If I moved an object in a circle, would I have done any work? (1 Viewer)
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Nailgun
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Suprisingly no-one has replied to you yet lel, I was going to but I thought it would be better for someone who legit knew what they were talking about lol
I googled a bit, and I think I sort of know the answer
Here are some references I found useful
https://www.physicsforums.com/threads/displacement-and-work.322461/
http://physics.stackexchange.com/questions/184659/work-force-x-distance-vs-displacement
https://en.wikipedia.org/wiki/Conservative_force
basically, I think that there is work done on the object because you pushing it is not a conservative force, so at each infitesmal portion of the path, you have changed the kinetic energy. I understand that work can be negative only with a conservative force so when you throw a ball up against gravity and catch it, the net work done in that system is zero, however with a non-conservative force like friction, it merely adds up at each portion of the journey (I think)
Take all this with 500 tons of salt though, not entirely sure that I am correct
I googled a bit, and I think I sort of know the answer
Here are some references I found useful
https://www.physicsforums.com/threads/displacement-and-work.322461/
http://physics.stackexchange.com/questions/184659/work-force-x-distance-vs-displacement
https://en.wikipedia.org/wiki/Conservative_force
basically, I think that there is work done on the object because you pushing it is not a conservative force, so at each infitesmal portion of the path, you have changed the kinetic energy. I understand that work can be negative only with a conservative force so when you throw a ball up against gravity and catch it, the net work done in that system is zero, however with a non-conservative force like friction, it merely adds up at each portion of the journey (I think)
Take all this with 500 tons of salt though, not entirely sure that I am correct
So I have a similar work related question
A 12kg car is moving south 10ms^-1 a force of 48 Newtons acting towards north over 5 seconds. Calculate:
a) the initial kinetic energy of the car (I got that)
b) the work done on the car by the force
So is the answer 0j (by doing final - initial kinetic energy, or force*displacement, where the displacement is 0), or is it 1200j ( by adding 600j of work done to stop it then another 600j of work to accelerate it in the opposite direction).
Half of my class and the teacher says its 1200j (I think this is what Nailgun said for the work in circle above) while the other half and the answers at the back of the worksheet says its 0j. so I don't really know what's the right one, greatly appreciate if anyone could help me out.
A 12kg car is moving south 10ms^-1 a force of 48 Newtons acting towards north over 5 seconds. Calculate:
a) the initial kinetic energy of the car (I got that)
b) the work done on the car by the force
So is the answer 0j (by doing final - initial kinetic energy, or force*displacement, where the displacement is 0), or is it 1200j ( by adding 600j of work done to stop it then another 600j of work to accelerate it in the opposite direction).
Half of my class and the teacher says its 1200j (I think this is what Nailgun said for the work in circle above) while the other half and the answers at the back of the worksheet says its 0j. so I don't really know what's the right one, greatly appreciate if anyone could help me out.
Drsoccerball
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I think it may be a more reasonable answer to say that since the circle experiences friction it has done work as friction is a non-conservative force.
EDIT: NVM didn't read your last line
...anyone?
Nailgun
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I'm pretty sure it's 1200, friction provides the centripetal force to turn, and hence since it is inherently non-conservative it should not be 0.
There's no reason that the work would be counteracted to have a net of zero
leehuan
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...
If it were the net energy of the entire car after the motion has ended THEN I'd argue it's 0J
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leehuan
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It's not centripetal motion this time.
Fizzy_Cyst
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I would say that the work would be equal to zero.
The Ek of the car at t = 0 and t = 5 would be identical, hence no work done.
Initially you would be doing negative work (force is in opp direction to its motion) then after 2.5s the work would be positive (and cancel out the negative work). This idea is the same as lifting an object up from the ground and then dropping it back to the ground.
The Ek of the car at t = 0 and t = 5 would be identical, hence no work done.
Initially you would be doing negative work (force is in opp direction to its motion) then after 2.5s the work would be positive (and cancel out the negative work). This idea is the same as lifting an object up from the ground and then dropping it back to the ground.
I'm pretty sure work is only done if there is a force component acting in the DIRECTION of displacement. Since the centripetal force is always perpendicular to the direction of displacement, there is no force acting in the direction of displacement and thus no work is done on an object in uniform circular motion. For the second question, I'm still a little unsure, but it should be what you said.