AP EAMCET202021 Sep 2020Morning ShiftPhysicsLaws of MotionActual
If (100 ~N ) force is applied to (10 ~kg ) block as shown in the diagram, the acceleration of (40 ~kg ) slab is
Options
- A(1.65 ~ms ⁻² )
- B(0.98 ~ms ⁻² )
- C(0.5 ~ms ⁻² )
- D(0.25 ~ms ⁻² )
Correct answer
B. (0.98 ~ms ⁻² )
Step-by-step solution
Static friction force between (10 ~kg ) and (40 ~kg ) block, ( aligned F_s & = _s R=0.6 m g & =0.6 10 9.8=58.8 ~N aligned ) Here, we see that the applied force ((F=100 ~N ) ) is greater than friction force, hence (10 ~kg ) block will start motion due to application of (100 ~N ) force. Due to motion, kinetic friction force ( aligned f_k & = _k R=0.4 mg & =0.4 10 9.8=39.2 ~N aligned ) (40 ~kg ) body experiences a force of (f_k=39.2 ~N ) ( ) Acceleration of (40 ~kg ) slab (a= f_k 40 = 39.2 40 =0.98 ~ms ⁻² )