Question

A 5.00-kg block is in contact on its right side with a 2.00-kg block. Both blocks rest on a horizontal frictionless surface. The 5.00-kg block is being pushed on its left side by a horizontal 20.0-N force. What is the magnitude of the force that the 5.00-kg block exerts on the 2.00-kg block?

Answer 1

Explanation:

The relation between force, mass and acceleration is as follows.

               F = ma

or,          a = $\frac{F}{m}$

As two blocks are in contact with each other. Hence, total mass will be as follows.

                mass = 5 kg + 2 kg

                          = 7 kg

Now, we will calculate the acceleration as follows.

             a = $\frac{F}{m}$

                = $\frac{20}{7}$

Hence, force exerted by mass of 2 kg on a mass of 5 kg will be calculated as follows.

               $20 - F_{1} = 5 \times \frac{20}{7}$

                   $F_{1}$ = 5.714 N

Thus, we can conclude that magnitude of the force that the 5.00-kg block exerts on the 2.00-kg block is 5.714 N.