Understanding Work Done


by Nubcake
Tags: work
Nubcake
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#1
Jan10-13, 12:29 PM
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I have always managed to solve problem asking how much work is being done but then once I start to think about the work being done I get confused. E.g A barrel of weight 200N is raised by a vertical distance of 1.8m by being moved along a ramp. The work done here would be 200 x 1.8 =360J . I do not understand why 200N is used since it is the weight acting vertically downwards opposite the distance moved in the direction of the force. So why is work being done shouldn't work be done when the barrel is moving in the same direction as its own weight? Or is it that 200N is the force that the person is giving to move it up along the ramp hence it is in the same direction of the force.Can someone clear this up for me?
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jedishrfu
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Jan10-13, 12:52 PM
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When you change heights you change the barrel's potential energy. By convention we arbitrarily use 0J potential energy at ground level and when you raise it 1.8 m then you get PE = F * d = 200N * 1.8m = 360 J then if you let it fall back to the ground it have 360 J of kinetic energy of movement at ground level which it then dissipates on impact.
Nubcake
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#3
Jan10-13, 01:04 PM
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Quote Quote by jedishrfu View Post
When you change heights you change the barrel's potential energy. By convention we arbitrarily use 0J potential energy at ground level and when you raise it 1.8 m then you get PE = F * d = 200N * 1.8m = 360 J then if you let it fall back to the ground it have 360 J of kinetic energy of movement at ground level which it then dissipates on impact.
I was more concerned about the direction and the force ; doesn't the weight of the barrel act in the opposite direction of movement?

cepheid
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Jan10-13, 01:37 PM
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Understanding Work Done


Quote Quote by Nubcake View Post
I was more concerned about the direction and the force ; doesn't the weight of the barrel act in the opposite direction of movement?
Yes. So the work done by gravity is negative (-360 J). By definition, the change in gravitational potential energy is equal to the negative of the work done by gravity. So, the change in potential energy is +360 J (potential energy increases when gravity does negative work).

The work done by the person pushing the thing up the ramp is positive, since the displacement and force are in the same direction. Let's assume that there is negligible friction. Let's also assume that the object is pushed up the ramp at a constant speed (neglecting the initial acceleration to get it moving). Then the forces parallel to the ramp have to be balanced, which means that the person pushes with a force equal to mgsinθ, which is the component of the weight that acts "down the ramp." The work done is then mgsinθ*d, where d is the distance travelled along the ramp. But from basic trigonometry, d = h/sinθ, so W = mgsinθ*(h/sinθ) = mgh.

The person does an amount of work equal to mgh = +360 J (but the ramp allows him to do so with a smaller force than if he just lifted it vertically).

Notice that the person does +360 J of work, and gravity does -360 J of work on the object, so the NET work done on the object is 0 (which makes sense because it has 0 net force on it). This explains why it does not gain any kinetic energy. (Recall that the work-energy theorem says that the work done on an object is equal to its change in kinetic energy).
Chestermiller
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#5
Jan10-13, 07:07 PM
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What cepheid is saying is that he uses less force to push it up the ramp, but he has to push it for a greater distance. So the product of the force times the distance comes out the same as if he had just lifted it vertically with a larger force over a smaller distance.


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