Question

Two blocks lie side by side on a frictionless table. The block on the left is of mass \(m\); the one on the right is of mass \(2 m .\) The block on the right is pushed to the left with a force of magnitude \(F,\) pushing the other block in turn. What force does the block on the left exert on the block to its right?

Short answer

Answer: The force exerted by the left block on the block to its right is \(\dfrac{F}{3}\).

Step-by-step solution

Define the System

Let's consider the whole system of the two blocks together.

Apply Newton's Second Law to the System

According to Newton's Second Law, the net force acting on the system is equal to the total mass of the system multiplied by its acceleration. The net force acting on the system is F and the total mass of the system is m + 2m = 3m. So, we can write: \(F = (3m)a\) Where F is the force exerted on the right block and a is the acceleration of the system.

Calculate the Acceleration of the System

We can calculate the acceleration a of the system by solving the equation from Step 2: \(a = \frac{F}{3m}\)

Apply Newton's Second Law to the Left Block

Now we will consider the left block separately. Let's denote the force exerted by the left block on the right block as \(F_1\). According to Newton's Second Law, \(F_1\) must be equal to the left block's mass multiplied by its acceleration: \(F_1 = m \cdot a\)

Calculate the Force Exerted by the Left Block

We can calculate the force exerted by the left block on the right block by substituting the acceleration of the system that we found in Step 3: \(F_1 = m \cdot \frac{F}{3m}\) Simplifying the equation and canceling the mass m: \(F_1 = \frac{F}{3}\) So the force exerted by the left block on the block to its right is \(\dfrac{F}{3}\).