A car with a mass of 1 ton, moving away, covered the first 50 m in 10 s.

A car with a mass of 1 ton, moving away, covered the first 50 m in 10 s. What work did the car engine do if the coefficient of resistance to movement is 0.05

m = 1 t = 1000 kg.

g = 10 m / s2.

V0 = 0 m / s.

S = 50 m.

t = 10 s.

μ = 0.05.

A -?

We express the work of the car engine A by the formula: A = Ft * S, where Ft is the traction force of the car, S is the movement of the car.

For the movement of a car Let us write 2 Newton’s law in vector form: m * a = Ft + m * g + N + Ftr, where Ft is the engine thrust force, m * g is the gravity force, N is the reaction force of the road surface, Ftr is the friction force …

ОХ: m * a = Fт – Fтр.

OU: 0 = – m * g + N.

Ft = m * a + Ftr.

N = m * g.

The friction force Ffr is expressed by the formula: Ffr = μ * N = μ * m * g.

Fт = m * a + μ * m * g = m * (a + μ * g).

S = V0 * t + a * t ^ 2/2 = a * t ^ 2/2.

a = 2 * S / t ^ 2.

Ft = m * (2 * S / t ^ 2 + μ * g).

A = m * (2 * S / t ^ 2 + μ * g) * S.

A = 1000 kg * (2 * 50 m / (10 s) 2 + 0.05 * 10 m / s2) * 50 m = 75000 J.

Answer: the car engine did the work A = 75000 J.