Question

1. A 9.32 x 10³N Car is traveling West on 14 towards the "Affle" house. Let the force onto
the wheels be set to be twice that of the car's total weight. The coefficient of kinetic
friction is said to be 0.222.
a. Determine the acceleration of the car
2. Determine the acceleration of the crate. A rightward force of 302 N is applied to an 18.6
kg crate to accelerate it across the floor. The coefficient of friction between the crate and
the floor is 0.750.

Answer 1

(1) The acceleration of the car is determined as 17.42 m/s².

(2) The acceleration of the crate is determined as 8.89 m/s².

Acceleration of the car

The acceleration of the car is calculated from the net force acting on the car.

∑F = ma

F - Ff = ma

F - μW = ma

where;

• F is the applied force on the car = 2 times weight
• W is weight of the car
• μ is coefficient of kinetic friction
• m is mass of the car

m = W/g

m = (9320)/(9.8)

m = 951.02 kg

2(9320) - 0.222(9320) = 951.02a

16,570.96 = 951.02a

a = 17.42 m/s²

Acceleration of the crate

F - μW = ma

F - μmg = ma

302 - 0.75(18.6 x 9.8) = 18.6a

165.29 = 18.6a

a = 8.89 m/s²

Thus, the acceleration of the car is determined as 17.42 m/s².

The acceleration of the crate is determined as 8.89 m/s².

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