KCET2024PhysicsLaws of Motion
A block of certain mass is placed on a rough inclined plane. The angle between the plane and the horizontal is 30^ . The coefficients of static and kinetic frictions between the block and the inclined plane are 0.6 and 0.5 respectively. Then, the magnitude of the acceleration of the block is [Take .g=10 ~ms ⁻² ]
Options
- A2 ~ms ⁻²
- Bzero
- C0.196 ~ms ⁻²
- D0.67 ~ms ⁻²
Correct answer
B. zero
Step-by-step solution
From the diagram, frictional force f_s= R= m g 30^ =0.6 m g 3 2 =0.3 3 mg =0.5196 mg Magnitude of force pulling the block along the plane downward, F^ =m g 30^ =m g / 2=0.5 mg Since, f_s F^ Hence, block will not move. a=0