A body weighing 0.2 kg is thrown upward at a speed of 6 m / s.

A body weighing 0.2 kg is thrown upward at a speed of 6 m / s. Determine the kinetic energy of the body and the maximum height to which it will rise.

m = 0.2 kg.

g = 9.8 m / s2.

V = 6 m / s.

Ek -?

hmax -?

The kinetic energy of the body, Ek, is determined by the formula: Ek = m * V ^ 2/2, where m is the mass of the body, V is the speed of the body.

Ek = 0.2 kg * (6 m / s) ^ 2/2 = 3.6 J.

According to the law of conservation of total mechanical energy, when the body moves upward, its kinetic energy Ek transforms into potential En. At the maximum ascent height hmax, the body has only potential energy En. At this altitude, all kinetic energy has passed into potential Ek = En.

En = m * g * hmax.

m * g * hmax = m * V ^ 2/2.

hmax = V ^ 2/2 * g.

hmax = (6 m / s) ^ 2/2 * 9.8 m / s2 = 1.83 m.

Answer: the maximum height of the body was hmax = 1.83 m, Ek = 3.6 J.