Question

A mass of 0.5 kg is just able to slide down the
slope of an inclined rough surface when the angle
of inclination is 60°. The minimum force necessary
to pull the mass up the incline along the line of
greatest slope is (g = 10 m/s2)
(A) 20.25N (B) 8.66 N (C) 100N (D) 1N​

Answer 1

first of all, find coefficient of friction,

at equilibrium,

component of weight along plane = frictional force

or, mgsin60° = u × mgcos60° , where u is coefficient of friction.

or, u = tan60° = √3

hence, we get coefficient of friction is √3.

now, The minimum force required to pull the mass up the inclined, Fnet = component of weight + frictional force

[ here you should notice that when we pull mass up the inclined plane , friction acts downward along plane that's why, net force = component of weight + friction ]

now, net force = mgsin60° + umgcos60°

here, m = 0.5 kg, g = 10m/s² and u = √3

so, net force = 0.5 × 10 (√3/2 + √3 × 1/2 )

= 5 × (√3)

= 5 × 1.732

= 8.66 N

hence, option (B) is correct.