A ball weighing 50 g was thrown vertically upward with an initial velocity of 10 m / s. The ball rose to a height of 4 m

A ball weighing 50 g was thrown vertically upward with an initial velocity of 10 m / s. The ball rose to a height of 4 m and fell back. The force of resistance to the movement of the ball is?

m = 50 g = 0.05 kg.

g = 9.8 m / s2.

V0 = 10 m / s.

h = 4 m.

Fsopr -?

Let us find the change in the total mechanical energy ΔE of the ball during lifting: ΔE = E0 – E.

At the moment of throwing the ball, the total mechanical energy E0 consists only of the kinetic energy Ek0: E0 = Ek0.

Ek0 = m * V02 / 2.

E0 = m * V02 / 2.

E0 = 0.05 kg * (10 m / s) 2/2 = 2.5 J.

The total energy of the ball E at the maximum height consists only of the potential Еп: Е = Еп.

En = m * g * h.

E = m * g * h.

E = 0.05 kg * 9.8 m / s2 * 4 m = 1.96 m.

The total mechanical energy of the ball decreased by ΔE = 2.5 J – 1.96 J = 0.54 J due to the work of the resistance force A.

A = ΔE.

A = Fcopr * h.

Fcopr = ΔE / h.

Fcopr = 0.54 J / 4 m = 0.135 N.

Answer: the force of resistance to the movement of the ball is Fcopr = 0.135 N.