Manipal MET2015PhysicsLaws of Motion
A block of mass m is lying on the edge having inclination angle = ⁻¹ ( 1 5 ) . Wedge is moving with a constant acceleration, a=2 ~ms ⁻² . The minimum value of coefficient of friction , so that m remains stationary with respect to wedge is
Options
- A2 a
- B5 12
- C1 5
- D2 5
Correct answer
B. 5 12
Step-by-step solution
FBD of m in frame of wedge N=m g ( )-m a ( ) Now, f= N=m a +m g aligned & = a +g g -a & = a+g g-a = 5 12 = 5 12 aligned