標題:
Work, energy and power
發問:
It seems to me when calculating the work done by force on an object which travels on a horizontal plane, the length of the plane is involved but not in the case of an inclined plane. Such case for example, the work done by an object along an inclined plane = increase in P.E. of block + work done against... 顯示更多 It seems to me when calculating the work done by force on an object which travels on a horizontal plane, the length of the plane is involved but not in the case of an inclined plane. Such case for example, the work done by an object along an inclined plane = increase in P.E. of block + work done against friction. Should I also add (length of plane x force done by object along that direct)? Which = Increase in PE + work on distance of plane + work on friction? And if such statement is untrue, then by the equation Work=increase in PE + work done against friction, in cases of horizontal movements, the Work only equals: work done against friction? If it is a smooth plane, then the answer of the question: A block of 10kg moves 5m on a frictionless plane is 0? Please explain detailedly because I am so confused as a F.4 student!
最佳解答:
In an inclined plane, the work done by the force pulling an object up the plane contributes to the gain in potnetial energy of the object. Let a be the angle of the inclinded plane m be the mass of the object L be the length travelled by the object along the plane. Assume the object is moving up the plane at constant speed by a force F, we have F = m.g.sin(a) + Ff, wherr Ff is the frictional force and g is the acceleration due to gravity The work done by the force W = F.L i.e. W = [mg.sin(a) + Ff].L = (mg.sin(a)).L + Ff.L ------------------ (1) The second term on the right is the work done against friction The first term on the right, which is the work done by the force on moving a distance L along the plane, can be written as, (mg.sin(a)).L = mg.(L.sin(a)) = mg.h where h = L.sin(a) and from geometry, it is the (vertical) height attained by the object when it travells a distance L along the plane. thus, we have, work done by force = increase in PE + work done against friction that is to say, the term [increase in PE] just represents the[work done by the force along the plane]. On a horizontal surface, angle a=0, thus equation (1) becomes work done by force W = Ff.L the result is obvious for an object travelling at constant speed. The applied force F is used to overcome friction. If the plane is smooth, Ff=0, then no force is needed to keep the object moving at constant speed. Hence, no work is done.
其他解答:
Work, energy and power
發問:
It seems to me when calculating the work done by force on an object which travels on a horizontal plane, the length of the plane is involved but not in the case of an inclined plane. Such case for example, the work done by an object along an inclined plane = increase in P.E. of block + work done against... 顯示更多 It seems to me when calculating the work done by force on an object which travels on a horizontal plane, the length of the plane is involved but not in the case of an inclined plane. Such case for example, the work done by an object along an inclined plane = increase in P.E. of block + work done against friction. Should I also add (length of plane x force done by object along that direct)? Which = Increase in PE + work on distance of plane + work on friction? And if such statement is untrue, then by the equation Work=increase in PE + work done against friction, in cases of horizontal movements, the Work only equals: work done against friction? If it is a smooth plane, then the answer of the question: A block of 10kg moves 5m on a frictionless plane is 0? Please explain detailedly because I am so confused as a F.4 student!
最佳解答:
In an inclined plane, the work done by the force pulling an object up the plane contributes to the gain in potnetial energy of the object. Let a be the angle of the inclinded plane m be the mass of the object L be the length travelled by the object along the plane. Assume the object is moving up the plane at constant speed by a force F, we have F = m.g.sin(a) + Ff, wherr Ff is the frictional force and g is the acceleration due to gravity The work done by the force W = F.L i.e. W = [mg.sin(a) + Ff].L = (mg.sin(a)).L + Ff.L ------------------ (1) The second term on the right is the work done against friction The first term on the right, which is the work done by the force on moving a distance L along the plane, can be written as, (mg.sin(a)).L = mg.(L.sin(a)) = mg.h where h = L.sin(a) and from geometry, it is the (vertical) height attained by the object when it travells a distance L along the plane. thus, we have, work done by force = increase in PE + work done against friction that is to say, the term [increase in PE] just represents the[work done by the force along the plane]. On a horizontal surface, angle a=0, thus equation (1) becomes work done by force W = Ff.L the result is obvious for an object travelling at constant speed. The applied force F is used to overcome friction. If the plane is smooth, Ff=0, then no force is needed to keep the object moving at constant speed. Hence, no work is done.
其他解答:
此文章來自奇摩知識+如有不便請留言告知
1. It is not true.It is because W=Fs is already equal to W=mgh.Look at the following working you will understand it clearly. W=Fs,W=mgsinxs,W=mg(h/s)s,W=mgh<<=PE sinx=h/s,subsitude it to the above equation. 2. Since W=mgh=Fs.Thus,the second statement is also wrong.The answer should contains Fs and work done against friction.In the simple case,F will be a net force.As a result,F major contain the applied force and the friction force.In this case,the friction against friction should be omissed.
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