To apply the law of conservation of energy to an object launched upward in Earth

Question

To apply the law of conservation of energy to an object launched upward in Earth's gravitational field. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from Earth. The energy transformations that take place involve the object's kinetic energy K = (1/2)mv^2 and its gravitational potential energy U = mgh. The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation K_i + U_i = K_f + U_f where "i" denotes the "initial" moment and T denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. Strictly speaking, it is not the ball that possesses potential energy; rather, it is the system "Earth-ball." Although we will often potential energy of an elevated object," it is useful to keep in mind that the energy, in fact, is associated with the interactions object. The potential energy of the object at the moment of launch is Usually, the zero level is chosen so as to make the relevant calculations simpler. In this case, it makes good sense to assu level, but this is not by any means the only choice! Using conservation of energy, find the maximum height h_max to which the object will rise. Express your answer in terms of v and the magnitude of the acceleration of gravity g.

Explanation / Answer

let v be the maximum velocity, so, when object is having maximum velocity it'll have only kinetic energy

and when the object is having maximum height it'll have only potential energy

by conservation of energy

initial energy = final energy

0.5 * m * v^2 = mgh_max

0.5 * v^2 = gh_max

h_max = v^2 / 2g

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