Sand is being dropped from a stationary dropper at a rate of 0.5 kgs–1 on a conveyor belt moving with a velocity of 5 ms–l. The power needed to keep the belt moving with the same velocity will be
- 1.25 W
- 2.5 W
- 6.25 W
- 12.5 W
The Correct Option is D
Solution and Explanation
The correct option is(D): 12.5 W.
\(\frac{dm}{dt}=0.5\,kg/s\)
\(v=5m/s\)
F=2.5
\(P=F+\overline{v}\)
=(2.5)(5)W=12.5 W
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Concepts Used:
Speed and Velocity
The rate at which an object covers a certain distance is commonly known as speed.
The rate at which an object changes position in a certain direction is called velocity.
Difference Between Speed and Velocity:
Read More: Difference Between Speed and Velocity