Answer:
[tex]a=0.6\ m/s^2[/tex]
Explanation:
The attached figure shows the whole description. Considering the applied force is 100 N.
The acceleration of both blocks A and B, [tex]a=0.8\ m/s^2[/tex]
Firstly calculating the mass m using the second law of motion as :
F = ma
m is the mass
[tex]m=\dfrac{F}{a}[/tex]
[tex]m=\dfrac{100\ N}{0.8\ m/s^2}[/tex]
m = 125 kg
It suddenly encounters a surface that supplies 25.0 N a friction, F' = 25 N
[tex](F-F')=ma[/tex]
[tex](100-25)=125\times a[/tex]
[tex]a=\dfrac{75}{125}=0.6\ m/s^2[/tex]
So, the new acceleration of the block is [tex]0.6\ m/s^2[/tex]. Hence, this is the required solution.
