AP EAMCET201822 Apr 2018Evening ShiftPhysicsLaws of MotionActual
A block of mass 10 ~kg is placed on a horizontal frictionless surface and is attached to a cord which passes over two light frictionless pulleys as shown in the figure. The hanging block tied to the other end of the cord is initially at rest 2 ~m above the horizontal floor.If the hanging block strikes the floor 2 ~s after the system is released, then weight of the hanging block is ....... (g=10 ~ms ⁻² )
Options
- A22.22 N
- B11.11 N
- C1.11 N
- D2.22 N
Correct answer
B. 11.11 N
Step-by-step solution
For hanging block, aligned & s=u t+ 1 2 a t^2[ u=0] 2= 1 2 a 2^2 & a=1 ~ms ⁻² aligned So, velocity of hanging block = velocity of 10 ~kg block after 2 ~s array ll & v=u+a t & v=0+1 2 or v=2 ~ms ⁻¹ array From work - energy theorem, KE of 10 ~kg block = Work done by gravity on hanging block = 1 2 10 2^2=m 10 2 m=1 ~kg Weight of hanging block =1 10=10 ~kg 11.1 lN