Question

A block of mass m (2kg) is placed on a rough plank. Now plank is slowly rotated in vertical plane. Variation of reaction force acting on
block with angle g is given in figure. Then :-​

Answer 1

Answer:

answer i write down

Explanation:

θ=30

o

;m=1 kg;g=9.8 m/s

2

Along plane of inclination force is

mg sinθ=0.5mg=4.9 N

Since downward force is greater than the force of static friction, so the kinetic friction will act.

Thus, maximum Friction force is μ

k

mgcosθ=1.96

3

N