A car weighs 2,000 kg. It moves along a road by applying a force on the road with a parallel component of 560 N. There are two passengers in the car, each weighing 55 kg. If the magnitude of the force of friction experienced by the car is 45 N, what is the acceleration of the car?

The acceleration of the car is 0.244 m/s²

Explanation:

Given;

mass of the car, m = 2,000 kg

applied parallel force on the car, F = 560 N

mass of the two passengers, m = 2 x 55 kg = 110 kg

frictional force on the car (reducing horizontal motion), Fk = 45 N

Weight of the car acting downward, W = 2,000 x 9.8 = 19,600 N

weight of the two passengers acting downward, w = 110 x 9.8 = 1078 N

The net horizontal force on the car is given by;

∑Fx = F - Fk

∑Fx = 560 - 45

∑Fx = 515 N

Apply Newton's second law of motion;

F = ma

where;

m is the total mass of the system = mass of car + mass of passengers

m = 2000 + 110 = 2,110 kg

a = ∑Fx / m

a = 515 / 2110

a = 0.244 m/s²

Therefore, the acceleration of the car is 0.244 m/s²

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