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 = [tex]\frac{F}{m+M} [/tex]
a = [tex]\frac{80}{1+11} [/tex]
a = [tex]\frac{80}{12} [/tex]
a = 6.67m[tex]s^{-2}[/tex]
Let's write the Equation of Motion of the heavier block
[tex]F_{1}[/tex] = F - [tex]F_{2}[/tex]
Ma = F - [tex]F_{2}[/tex]
force exerted by the lighter block on the heavier block is calculated as
[tex]F_{2}[/tex] = F - Ma
[tex]F_{2}[/tex] = 80 - (11 x 6.67)
[tex]F_{2}[/tex] = 6.63 N
