# Calculate the force on 2 kg block?

## Mar 2, 2017

$\vec{F} = \frac{20}{3} N \approx 6.7 N$

#### Explanation:

We will need to directly use Newton's second and third laws to solve this problem.

Newton's third law states, in summary, that that if an object A imparts a force on another object B, then object B imparts an equal and opposite force on object A. This is loosely referenced as "every action has an equal and opposite reaction."

These equal and opposite forces constitute Newton's third law pairs or "action/reaction pairs." Note that in order for two forces to be third law pairs, they must act on different objects. For example, the normal force and force of gravity may be equal and opposite in various situations, but they act on the same object and therefore do not constitute an NIII pair.

In this particular situation, the NIII pair consists of the force of the 1 kilogram block on the 2 kilogram block, and the force of the 2 kilogram block on the 1 kilogram block. These forces are equal in magnitude, but one acts in the negative direction while the other acts in the positive direction. I will define to the right as the positive direction.

Let's set up statements of the net force on each block (NII). We'll only consider the parallel forces, as there is no net force perpendicular (blocks move horizontally, not vertically). Additionally, we will assume a frictionless surface since it has been denoted as "smooth."

For the 1 kg block:

${\vec{F}}_{n e t} = \sum {\vec{F}}_{x} = {\vec{F}}_{p} - {\vec{F}}_{2 o n 1} = {m}_{1} \vec{a}$

Where ${\vec{F}}_{p}$ is the force of the push from the left and ${\vec{F}}_{2 o n 1}$ is the force of the 2 kg block on the 1 kg block.

For the 2 kg block:

${\vec{F}}_{n e t} = \sum {\vec{F}}_{x} = {\vec{F}}_{1 o n 2} = {m}_{2} \vec{a}$

The forces ${\vec{F}}_{2 o n 1}$ and ${\vec{F}}_{1 o n 2}$ are an NIII pair. Notice that they are opposite in direction, i.e. one is positive and one is negative. We know that they are equal in magnitude.

Before we can determine the force on the 2 kg block, we'll need to calculate the acceleration of the system. We know that the blocks will accelerate at the same rate, so we can add the two statements of the net force to solve for $\vec{a}$.

${\vec{F}}_{p} - {\vec{F}}_{2 o n 1} + {\vec{F}}_{1 o n 2} = {m}_{1} \vec{a} + {m}_{2} \vec{a}$

The NIII pair cancels.

$\implies {\vec{F}}_{p} = \vec{a} \left({m}_{1} + {m}_{2}\right)$

$\implies \vec{a} = \frac{{\vec{F}}_{p}}{{m}_{1} + {m}_{2}}$

$= \frac{10 N}{1 k g + 2 k g}$

$= \frac{10}{3} \frac{m}{s} ^ 2$

We can now use this value for the acceleration of the system to calculate ${\vec{F}}_{1 o n 2}$:

${\vec{F}}_{1 o n 2} = \left(2 k g\right) \left(\frac{10}{3} \frac{m}{s} ^ 2\right)$

$= \frac{20}{3} N$

This is $\approx 6.7 N$.

We can even check our answer in a couple ways. One way is by plugging this value in for ${\vec{F}}_{2 o n 1}$ and calculating the acceleration. If we get $10 \frac{m}{s} ^ 2$, it works out. We can also plug in $\frac{10}{3} \frac{m}{s} ^ 2$ for $\vec{a}$ and calculate ${\vec{F}}_{2 o n 1}$, knowing it should be equal to ${\vec{F}}_{1 o n 2}$.

$\vec{a} = \frac{{\vec{F}}_{p} - {\vec{F}}_{2 o n 1}}{m} _ 1$

$\vec{a} = \frac{10 N - \frac{20}{3} N}{1 k g}$

$\vec{a} = \frac{10}{3} \frac{m}{s} ^ 2$

OR:

${\vec{F}}_{2 o n 1} = {\vec{F}}_{p} - {m}_{1} \vec{a}$

$= \left(10 N - \left(1 k g\right) \left(\frac{10}{3} \frac{m}{s} ^ 2\right)\right)$

$= \frac{20}{3} N$