A constant force F / began to act on the trolley, the mass of which is m = 400 kg. Under its action

A constant force F / began to act on the trolley, the mass of which is m = 400 kg. Under its action, the trolley traveled a path S = 5.5 m and acquired a speed V = 2 m / s. On the way of movement, the coefficient of friction is f = 0.012. determine the work A of this force.

m = 400 kg.

g = 10 N / kg.

S = 5.5 m.

V0 = 0 m / s.

V = 2 m / s.

f = 0.012.

A -?

We express the work A of the acting force by the formula: A = F * S.

We will assume that the force F is directed horizontally along the movement of the body.

m * a = F – Ftr – 2 Newton’s law.

F = m * a + Ftr.

Since the trolley moves equally accelerated and from a state of rest, then a = (V ^ 2 – V0 ^ 2) / 2 * S = V ^ 2/2 * S.

Ftr = f * m * g.

F = m * V ^ 2/2 * S – f * m * g = m * (V ^ 2/2 * S – f * g).

F = 400 kg * ((2 m /) ^ 2/2 * 5.5 m – 0.012 * 10 N / kg) = 97.44 N.

A = 97.44 N * 5.5 m = 536 J.

Answer: when the trolley moves, the force performs work A = 536 J.