AP EAMCET202120 Aug 2021Evening ShiftPhysicsLaws of MotionActual
A 30 kg slab B rests on a frictionless floor as shown in the figure. A 10 kg block A rests on top of the slab- B. The coefficients of static and kinetic friction between the block A and the slab B are 0 . 60 and 0 . 40 respectively. When block-A is acted upon by a horizontal force of 100 N , as shown, find the resulting acceleration of the slab- B. g = 9 . 8 m s - 2
Options
- A0 . 98   m   s - 2
- B1 . 47   m   s - 2
- C1 . 52   m   s - 2
- D1 . 31   m   s - 2
Correct answer
D. 1 . 31   m   s - 2
Step-by-step solution
The maximum frictional force, μ s N = 0 . 6 × 10 × 9 . 8 = 58   N As the frictional force is less than the applied force, the block and slab won't move together. For the block, f = μ m g = 0 . 4 × 10 × 0 . 98 ⇒ f = 39 . 2   N The slab moves due to the kinetic friction. Resulting acceleration of the slab, a = f M = 39 . 2 30 = 1 . 31   m   s - 2