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Sand falls at $\dfrac{dm}{dt}$ onto a belt moving at speed $v$. The force needed to keep the belt moving is:
A$F=\dfrac{dm/dt}{v}$
B$F=\dfrac{v}{dm/dt}$
C$F=v\dfrac{dm}{dt}$
D$F=\dfrac{1}{2}v\dfrac{dm}{dt}$
Answer & Solution
Correct answer: C. $F=v\dfrac{dm}{dt}$
1. Newton's second law in its general form is $F=\dfrac{d(mv)}{dt}$, the rate of change of momentum.
2. The belt speed $v$ is constant, so only the mass on the belt changes.
3. That gives $F=v\dfrac{dm}{dt}$, equivalently $F=\dfrac{mv-mu}{t}$.
4. With $v=5$ cm/s $=0.05$ m/s and $\dfrac{dm}{dt}=100$ g/s $=0.1$ kg/s, the force is $0.005$ N.
5. Dividing the rate by the speed instead, which gave an answer of 20 N, was the recorded error and produces the wrong units.
_Source: NECTA ACSEE 2023 Physics 131/1, Question 4: Mechanics (Newton's Laws of Motion)_
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