Question
Download Solution PDFSand 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 :
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCONCEPT:
- 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
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