A child pulls on a toy locomotive of mass 0.979 kg with a force of 3.25 N to the right. The locomotive is connected to two train cars by cables. Friction in the axles results in an effective coefficient of kinetic friction between the floor and the train which is 0.110. One car has a mass of 0.952 kg and the other has a mass of 0.419 kg. What is the acceleration of the train?
What is the tension in the cable between the locomotive and the car connected to the locomotive?
Hints: For part (a), consider the locomotive and the 2 cars to be one “object”.
For part (b). consider the “object” to be the 2 cars.

(a) The acceleration of the train and cars is approximately 0.304 m/s²

(b) The tension between the locomotive and the car connected to the locomotive is approximately 0.4166 N

Explanation:

The given parameter are;

The mass of the toy locomotive = 0.979 kg

The mass of one car = 0.952 kg

The mass of the other car = 0.419 kg

The total mass of the train locomotive and cars = 0.979 + 0.952 + 0.419 = 2.35 kg

The weight of the train locomotive and cars = 9.81 m/s² × 2.35 kg = 23.0535 N

The normal force of the train and cars = The weight of the train locomotive and cars = 23.0535 N

The coefficient of dynamic friction = The normal force × The coefficient of dynamic friction

The coefficient of dynamic friction = 23.0535 × 0.110 = 2.535885 N

The applied force = 3.25

The net force acting on the train locomotive and cars = 3.25 - 2.535885 0.714115 N

(a) Force = Mass × Acceleration

Acceleration = Force/Mass

∴ The acceleration of the train and cars = 0.714115 N/(2.35 kg) ≈ 0.304 m/s²

(b) The tension between the locomotive and the car connected to the locomotive = T₂ = ( (m₁ + m₂)/ (m₁ + m₂ + m₃)) × F

Therefore, T₂ =  ( (0.952 + 0.419)/ (0.952 + 0.419 + 0.979)) × 0.714115 ≈ 0.4166 N.

1/0.14nm = 10^10/1.4 a lot