physics qn
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hi guys, i would like to know based on the logic of work done, am i right to say that if a person of weight 700N is to run 500m, the amount of work done is 35000J? but den again, i was tinking since gravitational force (weight) is downwards and distance ran is forward, there is no work done at all? or is there some other force that nids to be accounted for to calculate work done?
2nd qn, if indeed the calculation above is true, am i right to say that a heavier person will do more work provided both subjects ran the same distance?
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Hello, what level is your question at? Is it an O level question or university question?
My knowledge is only up to o level, thus answer/theory may be wrong, better check with your teacher.
You are right to say that the work done is 35000J, if the amount of force needed to move a 700N person is exactly 700N(theorily it's impossible), and assuming frictional, gravitational force, and other external forces are negligible. Also, the direction in which you are moving, should be in the same direction of the force.
Newton's first law states that every object continues in its state of rest, or of uniform motion in a straight line, unless compelled to change that state by external forces acted upon it.An object that is at rest will stay at rest unless an unbalanced force acts upon it.
Now base on newton's first law, if a person is 700N in weight, a balanced force of 700N will not be enough to move the person(assuming all other force are negligible), and if there are no movement, distance = 0, and Workdone = Force x 0, theorily workdone = 0.
Now if i were to take gravitional force into account, i believe i need to draw a vector diagram of the forces acting upon the person.
Since there are two forces, you cannot use the usual Workdone = F x Distance, you need to use Workdone = Force x Distance x Cos θ.
After finding the net force, you need to take into account the frictional force.
Also, you need to take into consideration whether the person is accelerating or not.
I believe a heavier person will require more work because you need a greater unbalanced force when compared to a lighter person to move him. -
its juz a random qn i tot abt. coz i was tinking if weight (gravitational force) and distance run is not in the same direction and hence the work done eqn cannot be used in this instance, how is it that a heavier person does more work when he exercises the same distance and intensity? or are there other forces we should plug into the eqn?
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Because distance moved and gravitational force are not in the same direction,
the work done by the gravitational force is zero. Work done by you against gravitational forces is also zero.
A heavier person may do more work because you will have to consider human physiology who is not in the physics syllabus. When you run, you have motions of up and down, thus you do work against gravity. Furthermore, when you are heavier, the normal reaction force by the ground on you is larger, and hence the maximum frictional force is larger. Friction acts in the same direction as distance moved, hence your work done is also larger.
If the above paragraph stumps you, you can forget about it, because it is out of O level syllabus.
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Frictional force must be considered to solve this question. Frictional force are determined by both gravitional force and the frictional coefficient, so the work done wont be definitely larger for a heavier person. Try to think, if you seat in the car and travel a long journey, any work done?
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Originally posted by aspiration:
Frictional force must be considered to solve this question. Frictional force are determined by both gravitional force and the frictional coefficient, so the work done wont be definitely larger for a heavier person. Try to think, if you seat in the car and travel a long journey, any work done?
Frictional force is not determined by gravitational force.
It is determined by the coefficient of friction as you mentioned, as well as the normal reaction force between the surfaces in contact.
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Yes, you are right. But in this case, these two are same.
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Originally posted by eagle:
Frictional force is not determined by gravitational force.
It is determined by the coefficient of friction as you mentioned, as well as the normal reaction force between the surfaces in contact.
U can explain by saying a heavier person will have a bigger surface area, thus more resistance...as for gravitional force, it will be constant thruout... thus cannot use it to compare... whether a lighter person or a heavier person, the gravitional force will be the same... Reference to Galileo weight dropping experiment at the leaning tower of Pisa...
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Originally posted by ^Acid^ aka s|aO^eH~:
U can explain by saying a heavier person will have a bigger surface area, thus more resistance...as for gravitional force, it will be constant thruout... thus cannot use it to compare... whether a lighter person or a heavier person, the gravitional force will be the same... Reference to Galileo weight dropping experiment at the leaning tower of Pisa...
Wrong, friction does not depends on the size of the surface area at all.
For a stationary object that is of the same weight, the normal reaction force will remain the same no matter the surface area. You can have different sizes of surface area but your amount of friction stays the same, as long as the normal reaction force between the two surfaces stays the same.
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Originally posted by aspiration:
Yes, you are right. But in this case, these two are same.
It could be different. It will only be the same if we consider the person to have zero vertical acceleration, which in real life may not necessarily be so.
For the context of simplifying the question, yes, the person will have zero vertical acceleration when moving horizontally.
While normal reaction force and gravitational force are of the same magnitude above due to zero vertical acceleration (which in turn implies zero resultant force due to Newton's 2nd law), it is still wrong to say frictional force depends on the gravitational force, because it doesn't. Normal reaction force and gravitational force are two forces of different nature, and it does not make much sense to say that they are the same.
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Originally posted by ultimatenolifer:
its juz a random qn i tot abt. coz i was tinking if weight (gravitational force) and distance run is not in the same direction and hence the work done eqn cannot be used in this instance, how is it that a heavier person does more work when he exercises the same distance and intensity? or are there other forces we should plug into the eqn?
I understand work done as a state property. Therefore, if one looks at the final state and the energy expense versus the initial state, it would be breaking physical laws that the final energy state is higher than the initial energy state assuming mass is moved.
Assuming acceleration is synonomous with your statement of intensity then it is obvious that the heavier the mass the more work is done.
An interesting case is when the mass tends to infinity and the acceleration tends to zero, i.e. a massive slow-moving object. This came up whilse thinking about your question, which is intuitively interesting.
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Hi! Hope I'm not too late for this question.
Now let's understand work done. It is the force multiplied by the distance in the direction of the force.
For your case of moving a object which weighs 700N for 500m, you have to see if there is a force first.
So, I'm not sure if you are aware of this thing, there is something called a work-energy thereom, which states that the work done on an object is equal to the change in kinetic of an object.
Or by Newton's second law, F=ma, if you change your velocity, there is a force.
So back to your question of pushing an object over a horizontal distance, depending on the change of speed, you can get 0 work done or a positive work done..
Now let's hope I didn't bore anyone, and hopefully my theories are right...But it is madness to say we did not work if we run at a constant speed!
Let's now consider a person running. What forces must act on the body to speed him up. The only, neglecting air resistance for simplicity, is the force of the feet on the ground. One pushes the ground backwards while the ground pushes you forward. So there is a force acting over a distance=work done
But in a constant speed, we will slow down, due to heat dissipated etc. Hence a force must keep acting on us to move us.
Let's get back to the question of the magnitude of work done. Well it isn't really that simple as 500x700. Because by F=ma, any force will accelerate you.
So then what is the magnitude of the force then? Well there are two types of friction, one is the one that we commonly associate with, the one that resist movement, and another one called static friction. The static friction is proportional to the weight of the body. It is a force that resist the start of motion. Think about it, when you run, your feet do not move as you push yourself forward, right?
So it is true to say heavier people do more work? erm debatable
Ok i know i'm simplifying lots of things, but that's cause my knowledge is limiting me. For example even if the static friction provides the force, the object still didn't move when the force is applied. But i'm guessing is the work in creating a moment. But oh wellz...
And finally is there work to overcome gravity? Yes since you gain gravitational potential energy when you bounce up, by conservation of enery, you must have did wrok to get there. But unless you're pole vaulting throughout the run, the work done should be neglible.....
Once again I only know so little. So I may be critically wrong...
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Originally posted by LoveLoveTT:
Hi! Hope I'm not too late for this question.
Now let's understand work done. It is the force multiplied by the distance in the direction of the force.
For your case of moving a object which weighs 700N for 500m, you have to see if there is a force first.
So, I'm not sure if you are aware of this thing, there is something called a work-energy thereom, which states that the work done on an object is equal to the change in kinetic of an object.
Or by Newton's second law, F=ma, if you change your velocity, there is a force.
So back to your question of pushing an object over a horizontal distance, depending on the change of speed, you can get 0 work done or a positive work done..
Now let's hope I didn't bore anyone, and hopefully my theories are right...But it is madness to say we did not work if we run at a constant speed!
Let's now consider a person running. What forces must act on the body to speed him up. The only, neglecting air resistance for simplicity, is the force of the feet on the ground. One pushes the ground backwards while the ground pushes you forward. So there is a force acting over a distance=work done
But in a constant speed, we will slow down, due to heat dissipated etc. Hence a force must keep acting on us to move us.
Let's get back to the question of the magnitude of work done. Well it isn't really that simple as 500x700. Because by F=ma, any force will accelerate you.
So then what is the magnitude of the force then? Well there are two types of friction, one is the one that we commonly associate with, the one that resist movement, and another one called static friction. The static friction is proportional to the weight of the body. It is a force that resist the start of motion. Think about it, when you run, your feet do not move as you push yourself forward, right?
So it is true to say heavier people do more work? erm debatable
Ok i know i'm simplifying lots of things, but that's cause my knowledge is limiting me. For example even if the static friction provides the force, the object still didn't move when the force is applied. But i'm guessing is the work in creating a moment. But oh wellz...
And finally is there work to overcome gravity? Yes since you gain gravitational potential energy when you bounce up, by conservation of enery, you must have did wrok to get there. But unless you're pole vaulting throughout the run, the work done should be neglible.....
Once again I only know so little. So I may be critically wrong...
As we said in college, full of fluff, talking through your nose.
Make a conclusion, man/woman.