# A constant horizontal force of 20 N is applied to block A (4.0 kg) which is pushed against block...

## Question:

A constant horizontal force of 20 N is applied to block A (4.0 kg) which is pushed against block B (6.0 kg). The blocks slide over a surface, along the x-axis. What is the force from block A on block B when friction is neglected?

## Force:

Force is the measure of the amount of acceleration a mass is experiencing. There are two types of forces contact and non-contact forces. An example of a contact force is a pull or a push whereas gravity is an example of a non-contact force.

## Answer and Explanation:

**Given:**

{eq}\begin{align*} F &= \ N \\ m_1 &= 4\ kg \\ m_2 &= 6\ kg \\ \mu &= 0.1 \\ g &= \ m/s^2 \\ \text{Total mass}, m_t &= 10\ kg \\ \text{Frictinal force}, f_k &= \mu mg = \ N \\ \end{align*} {/eq}

According to the above figure,

{eq}\begin{align*} \text{Net force}, F_{net} &= 10 \ N \\ \because F &= ma \\ \therefore a_{net} &= \frac{f_{net}}{m_t} = 1 \ m/s^2 \\ \end{align*} {/eq}

In the above figure, the 4N force is calculated with the formula of mass F= ma and the other 4N is due to frictional force as the calculated force is already included in the 20N force that is why it is getting subtracted and frictional force is opposing force so it will also get subtracted from 20N.

{eq}\begin{align*} \text{contact force on the block A} &= F- 4 -4 = 12\ N \\ \end{align*} {/eq}

Since the contact force experienced by block A is 12N so the force from block A on block B will also be 12N.

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