# After turning off the engine, a car weighing 20kN stops under the influence of a friction force of 4kN

**After turning off the engine, a car weighing 20kN stops under the influence of a friction force of 4kN, passing 20m along a horizontal road. We need to calculate the work**

P = 20 kN = 20,000 N.

Ftr = 4 kN = 4000 N.

S = 20 m.

A -?

Mechanical work A of the friction force Ftr is determined by the formula: A = Ftr * S * cosα, Ftr is the friction force that acts on the body, S is the movement of the body, ∠α is the angle between the direction of action of the force and the displacement.

Friction force Ffr directed in the opposite direction to displacement S, ∠α = 180, cos180 = -1.

A = – Ftr * S.

A = – 4000 N * 20 m = – 80,000 J.

The force of gravity is directed perpendicular to the displacement, ∠α = 90, cos90 = 0.

Therefore, the work of gravity is 0.

Answer: the work of the friction force A = – 80,000 J.