Question

Consider the following system, where F = 80 N, m = 1 kg, and M = 11 kg M m F What is the magnitude of the force with which one block acts on the other?

Answer 1

Answer:

Force exerted by the lighter block on the heavier block is 6.63 N

Explanation:

Given Data

F = 80N

m = 1kg

M = 11kg

Solution:

*We assume that there is no friction

Calculating the acceleration of the system

a = $\frac{F}{m+M}$

a = $\frac{80}{1+11}$

a = $\frac{80}{12}$

a = 6.67m$s^{-2}$

Let's write the Equation of Motion of the heavier block  

$F_{1}$ = F - $F_{2}$

Ma = F - $F_{2}$

force exerted by the lighter block on the heavier block is calculated as

 $F_{2}$ = F - Ma

 $F_{2}$ = 80 - (11 x 6.67)  

 $F_{2}$ = 6.63 N