A car with a mass of 2 tons rises up a hill, the slope of which is 0.2. On a distance of 32 m, the speed of the car increased

A car with a mass of 2 tons rises up a hill, the slope of which is 0.2. On a distance of 32 m, the speed of the car increased from 21.6 km / h to 36 km / h. Assuming that the movement of the car is uniformly accelerated, find the thrust of the engine. The coefficient of resistance to movement is 0.02.

m = 2 t = 2000 kg.

g = 10 m / s2.

S = 32 m.

V0 = 21.6 km / h = 6 m / s.

V = 36 km / h = 10 m / s.

h / S = sinα = 0.2.

μ = 0.02.

Fт -?

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

ОХ: m * a = Fт – Fтр – m * g * sinα.

OU: 0 = – m * g * cosα + N.

Ft = m * a + Ftr + m * g * sinα.

N = m * g * cosα.

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

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

We express the acceleration of the car by the formula: a = (V ^ 2 – V0 ^ 2) / 2 * S.

a = ((10 m / s) ^ 2 – (6 m / s) ^ 2) / 2 * 32 m = 1 m / s2.

cosα = √ (1 – sin2α) = √ (1 – (0.2) ^ 2) = 0.98.

Ft = 2000 kg * (1 m / s2 + 0.02 * 10 m / s2 * 0.98 + 10 m / s2 * 0.02) = 2800 N.

Answer: when driving on an inclined plane, the traction force of the car engine is Ft = 2800 N.