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Chapter 4: Problem 86

A 2.0 -kg toy locomotive is pulling a 1.0 -kg caboose. The frictional force of the track on the caboose is \(0.50 \mathrm{N}\) backward along the track. If the train's acceleration forward is \(3.0 \mathrm{m} / \mathrm{s}^{2},\) what is the magnitude of the force exerted by the locomotive on the caboose?

Short Answer

Expert verified
Answer: 3.50 N

Step by step solution

01

Identify the forces acting on the caboose

There are two forces acting on the caboose: the force exerted by the locomotive, which we'll represent as \(F_{L}\), and the frictional force acting in the opposite direction, which is 0.50 N. We will denote the frictional force as \(F_{f}\).
02

Write the equation using Newton's second law

Newton's second law states that the net force acting on an object is equal to the product of its mass and acceleration: \(F_{net} = m \times a\). We can write the net force acting on the caboose as the difference between the force exerted by the locomotive and the frictional force: \(F_{net} = F_{L} - F_{f}\).
03

Substitute the given values and solve for the unknown force

We are given the mass of the caboose (m = 1.0 kg) and the train's acceleration (a = 3.0 m/s虏). We also know that the frictional force is \(F_{f} = 0.50\,\mathrm{N}\). Substituting these values into the equation from Step 2, we get: \(F_{net} = F_{L} - F_{f}\) \(m \times a = F_{L} - 0.50\,\mathrm{N}\) \(1.0\,\mathrm{kg} \times 3.0\,\mathrm{m/s^2} = F_{L} - 0.50\,\mathrm{N}\) \(3.0\,\mathrm{N} = F_{L} - 0.50\,\mathrm{N}\) Now, we will solve for \(F_{L}\): \(F_{L} = 3.0\,\mathrm{N} + 0.50\,\mathrm{N}\) \(F_{L} = 3.50\,\mathrm{N}\)
04

State the final answer

The magnitude of the force exerted by the locomotive on the caboose is \(3.50\,\mathrm{N}\).

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