Question

Block A is accelerating with Block B at a rate of 0.800 m/s2 along a frictionless surface. It suddenly encounters a surface that supplies 25.0 N a friction. What is the new Acceleration?

A. 0.4 m/s2

B. 1.0 m/s2

C. 0.8 m/s2

D. 0.6 m/s2

Answer 1

`a=0.6\ m/s^2`

Step-by-step explanation:

The attached figure shows the whole description. Considering the applied force is 100 N.

The acceleration of both blocks A and B,
`a=0.8\ m/s^2`

Firstly calculating the mass m using the second law of motion as :

F = ma

m is the mass

`m=(F)/(a)`

`m=(100\ N)/(0.8\ m/s^2)`

m = 125 kg

It suddenly encounters a surface that supplies 25.0 N a friction, F' = 25 N

`(F-F')=ma`

`(100-25)=125* a`

`a=(75)/(125)=0.6\ m/s^2`

So, the new acceleration of the block is
`0.6\ m/s^2`. Hence, this is the required solution.