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

Step-by-step explanation:

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

F = ma

or, a =
`(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 =
`(F)/(m)`

=
`(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 * (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.