1

Sand runs from a hopper at a constant rate $\frac{dm}{dt}$ onto a horizontal conveyer belt driven at a constant speed $v$ by a motor. The power needed to drive the belt can be calculated as follows:

Momentum change of sand mass $dm$ is $dp=vdm$. The force $F$ on the belt is $$F=\frac{dp}{dt}=v\frac{dm}{dt}$$ The power $P$ to drive the belt is $$P=Fv=v^2\frac{dm}{dt}$$ Now, let us calculate the rate of change of kinetic energy of the sand: $$\frac{dK}{dt}=\frac{1}{2}\frac{dm}{dt}v^2$$ Half the power given to the belt goes to giving kinetic energy energy to the sand. My question is, where does the other half go?

t2m
  • 378
  • 1
  • 8
user231188
  • 97
  • If you assume a parcel of sand with mass $\delta m$ accelerates at a constant rate from $v(0)=0$ to $v(\delta t)=v$ (i.e. $v(t)=v\frac{t}{\delta t}$), provided by the requisite force dictated by Newton's 2nd law $F=\frac{\delta m}{\delta t} v$, then the integrated power $\int F(t) v(t) dt = \int \left( \frac{\delta m}{\delta t} v\right)\left(v\frac{t}{\delta t} \right) dt = \frac{1}{2}\frac{\delta m}{\delta t}v^2 \delta t$, which is exactly the change in kinetic energy you stated. The error you made was in assuming that the instantaneous power was a constant $P=Fv$. – Arturo don Juan Sep 26 '22 at 16:40
  • @ArturodonJuan I don't see an error in it. In fact, thats how it is done in the solution manual of kleppner & kolenkow – user231188 Sep 26 '22 at 17:19
  • You are asking about the change in kinetic energy of individual, infinitesimal parcels of sand. The initial velocity when it is dropped onto the conveyor belt is zero, and final velocity is $v$. This shows that the velocity of the sand parcels is clearly not constant. I am saying that you need to apply this same logic to the calculation of the power, imparted by the conveyor belt onto the sand. – Arturo don Juan Sep 26 '22 at 20:06

0 Answers0