I want to know that, if we suppose that, if a ball is falling freely then work would be done on the ball by gravity so energy would be given to the ball? and when the ball rebounds after striking the ground work would be done by the ball against gravity but since the ball is working on itself so it’s energy would remains the same (ignoring dissipation of energy)?
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Work is done by a force applied to the object:
In physics, work is the energy transferred to or from an object via the application of force along a displacement.
Thus, when the ball is falling, the work is done on the ball by the gravitational force (while the ball is doing work on Earth, by the force applied to the latter.) When the ball bounces, the work is done on the ball by the ground (and the ball does work on the ground, since the forces come in pairs, applied to the interacting bodies - Newton's 3rd law.) After the ball has rebounded, there is again gravity doing work on the ball, slowing its ascent.
In the discussion above the ball is treated as a point-like object, which is the approximation that we usually use when applying Newton's law (which are formulated for such point-like objects.) However, a ball is a complex object, so we could consider also the dynamics of its inner parts (e.g., treating them in term as point-like objects described by the Newton laws.) In this case these internal parts interact via forces, and thus perform work on each other.