Discussion
A belt conveyor \( 60 \) m long carries \( 150 \) t/h against an effective friction force of \( 1.8 \) kN at \( 1.5 \) m/s. The power required is
Answer: A
Correct answer: 2.7 kW
Power is force times velocity:
\[ P = F v = 1800 \times 1.5 = 2700\ \mathrm{W} = 2.7\ \mathrm{kW} \]
Neither the length nor the tonnage enters directly - both are already accounted for in the effective friction force, which is what the belt drive must overcome. For a horizontal run this friction is the whole of the duty; an inclined belt would add the lifting power \( mgh/t \) on top.
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