# Boxes A and B are in contact on a horizontal friction-less surface. Box A has mass 20 kg and box B has mass 5kg.A horizontal force of 250N is exerted on box A. What is the magnitude of the force that box A exerts on box B?

This problem aims to familiarize us with a friction-less motion between two masses as a single system. The concept required to solve this problem includes acceleration, newtons law of motion, and the law of conservation of momentum.

In this particular problem, we require the help of newton’s second law, which is a quantitative definition of the transformations that a force can have on the motion of a body. In other words, it is the rate of change of the momentum of a body. This momentum of a body is equivalent to mass times its velocity.

For a body having a constant mass $m$, Newton’s second law can be composed in the form $F = ma$. If there are multiple forces acting on the body, it is equally accelerated by the equation. Contrarily, if a body doesn’t accelerate, no kind of force is acting on it.

The force $F = 250 \space N$ is causing acceleration to both the boxes.

Applying Newton’s second law to obtain the acceleration of the whole system:

$F = (m_A+ m_B)a_x$

Making $a_x$ the subject of the equation.

$a_x = \dfrac{F}{(m_A+m_B)}$

$a_x = \dfrac{(250)}{20+5}$

$a_x = 10 \space m/s^2$

As Box A is exerting force on Box B, both the boxes are accelerating at the same speed. So it can be said the acceleration of the whole system is $10\space m/s^2$.

Now applying the Newtons second law on box B and calculating the force $F$:

$F_A = m_ba_x$

$= 5 \times 10$

$F_A = 50 \space N$

Box A exerts the force of magnitude $50 \space N$ on box B.

## Example

Boxes A and B and C are in contact on a horizontal, frictionless surface. Box A has mass $20.0 kg$, box B has mass $5.0 kg$ and box C has a mass $15.0 kg$. A horizontal force of $200 N$ is exerted on box A. What is the magnitude of the force that box B exerts on box C and the Box A exerts on Box B?

The force $F = 200\space N$ is causing acceleration to all the boxes.

Applying Newton’s second law to acquire the acceleration of the entire system:

$F = (m_A+m_B+m_C) a_x$

Making $a_x$ the subject of the equation.

$a_x = \dfrac{F}{(m_A+m_B+m_C)}$

$a_x = \dfrac{(200)}{20 +5+15}$

$a_x = 5\space m/s^2$

As Box A is exerting force on Box B, and then Box B is exerting force on Box C, All the boxes are accelerating at the same speed. So it can be said the acceleration of the whole system is $5\space m/s^2$.

Now applying the Newtons second law on box C and calculating the force $F_B$.

$F_B = m_Ca_x$

$= 15 \times 5$

$F_B = 75 \space N$

Box B exerts the force of $75 \space N$ on Box C.

Now,

$F_A = m_Ba_x$

$= 5 \times 5$

$F_A = 25 \space N$

Box A exerts the force of $25 \space N$ on Box B.