Question

If 100N force is applied to a 10 kg block as shown in the diagram , then acceleration produced in the block


A. \[1.65{\text{ }}m/{s^2}\]
B. \[0.98\,m/{s^2}\]
C. \[1.2\,m/{s^2}\]
D. \[0.25\,m/{s^2}\]

Answer 1

Hint We first need to check if 100N of force is sufficient to move the top block, if it is, then the block will accelerate in the direction of the force, else there will be relative motion between the blocks. The force acting on the lower block will be the force due to limiting friction between the 10kg and 40kg block.

Complete step by step solution
100N force is acting on the top block which is 10kg. the limiting friction to move this block will
 \[f\, = \,\mu 10g\]
 \[f\, = \,0.6\,\times\,10\,\times\,9.81\]
 \[f\, = \,58.6{\text{ }}N\]
As \[100 > {\text{ }}58.6\] , the upper block will move with the force, and the lower block will move in the opposite direction
Net forces acting on the lower block is equal to 58.6N in the opposite direction, therefore the acceleration of the lower block is:

 \[
  58.6\, = \,40a \\
  a\, = \,\dfrac{{58.6}}{{40}} \\
  a{\text{ }} = {\text{ }}1.46 \\
 \]

Therefore no option is correct in this question

Note If there was friction between the lower block and ground, the net force would be the difference of the force due to the upper block and force of friction due to the lower block.