A car weighing 14 kN starts to move with an acceleration of 0.7 m / s2. The driving resistance force is 0.02

A car weighing 14 kN starts to move with an acceleration of 0.7 m / s2. The driving resistance force is 0.02 of the vehicle weight. Determine: a) the mass of the car; b) the force of gravity acting on the vehicle; c) thrust force developed by the engine; d) the speed and movement of the vehicle 10 s after the start of movement.

P = 14 kN = 14000 N.

V0 = 0 m / s.

a = 0.7 m / s ^ 2.

t = 10 s.

Fcopr = 0.02 * P.

g = 9.8 m / s ^ 2.

m -?

Fт -?

Ftraction -?

S -?

V -?

The force of gravity Ft and the weight of the body P is expressed by the formula: Ft = P = m * g.

Ft = 14000 N.

m = P / g.

m = 14000 N / 9.8 m / s ^ 2 = 1428.6 kg.

V = V0 + a * t.

V = 0.7 m / s ^ 2 * 10 s = 7 m / s.

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

S = 0.7 m / s ^ 2 * (10 s) ^ 2/2 = 35 m.

Let’s write 2 Newton’s law in vector form: m * a = Fthrust + Ft + N + Fcopr.

ОХ: m * a = Fthrust – Fcomp.

ОУ: 0 = Fт – N.

N = Fт.

m * a = F thrust – 0.8 * m * g.

m * a + 0.8 * m * g = F draft.

Fthrust = 1428.6 kg * (0.7 m / s ^ 2 + 9.8 m / s ^ 2) = 15000 N.

Answer: m = 1428.6 kg, Ft = 14000 N, Fthrust = 15000 N, V = 7 m / s, S = 35 m.