Sand falls vertically on a conveyor belt at a work rate of 0⋅1 kg/s. In order to keep the belt moving at a uniform speed of 2 m/s, the force required to be applied on the belt is :

CONCEPT:

• Force (F): The rate of change of momentum is called force.

​F ∝ dp/dt

• Acceleration (a): The rate of change of velocity is called acceleration.

it is  given by:

a = dv/dt

• Newton's third law of motion
  • Every action has an equal and opposite reaction.

CALCULATION:

​Fext  =  dp/dt  =  d(mv)/dt

m = mass of system as conveyor belt with sand drops at time t, v = velocity of conveyor belt at time t.

using product rule of differentiation

Fext = m(dv/dt) + v(dm/dt)

but v = constant, so dv/dt = 0

Fext = v(dm/dt) = Mv

Fext = is in direction of v̅  of belt.

since m.dv/dt = Fext + Freaction  ...................(i)

consider belt as a system with variable mass  dv/dt = 0 , 

m = mass of system at time t

Freaction  = vrec dm/dt  =  -v(dm/dt)

Fext  = v. dm/dt 

Fext  = 2m/s × 0.1 kg/s = 0.2 N