Power needed to keep conveyor belt running

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Taking change in the kinetic energy per unit time as power,

P = $\frac {d W}{dt} = \frac {d K}{dt} = \frac 1 2 \frac { dm} {dt} v^2 = 36.75$ watt...(1)

$P = \vec F \cdot \vec v = \frac{ dm} {dt} \vec v \cdot \vec v = \frac { dm} {dt} v^2 = 73.5$ watt...(2)

$dK = \vec F \cdot d \vec r = v^2 dm$ ...(3)

change in the kinetic energy per unit time = $\frac 1 2 (m + 1.5) v^2 - \frac 1 2 m v^2 = 1.5 ~ \frac 1 2 v^2$ ...(4)

I am getting two different answers.
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[Altair Tans]

Every sand particle doesn't start moving as soon as it falls...but over a short interval it has to overcome friction and attain the velocity...out of 73.5 watt 36.75 goes into kinetic energy and remaining half goes into work against friction

 

[Pushoam]

Let's say that the system is the sand particles upon the conveyor belt and the conveyor belt .

At time t, the mass of the system is m(t).

Now, the kinetic energy of the system at time t is $\frac 1 2 m(t) v^2$.

the kinetic energy of the system at time t + $dt$ is $\frac 1 2 (m(t) + dm)v^2$.

Now, the change in the kinetic energy of the system per unit time is $\frac 1 2 \frac { dm} {dt} v^2$. I am supposed to provide this much energy to the system per unit time. So, the required power is $\frac 1 2 \frac { dm} {dt} v^2$ = 36.75 W.

Now, the friction force is an internal force. Why should this be considered?

 

[Altair Tans]

Frictional force is internal...but there is always heat loss associated with it...you can't ignore thatMomentum change of the system due to friction will be zero owing to its internal nature but heat loss would have to be considered

 

[Pushoam]

What is happening is : when the sand particle falls on the conveyor belt, friction force works upon it till its speed becomes equal to the speed of conveyor belt. Work done by the friction force upon the particle is $\frac 1 2 1.5 v^2$ .

Now, work done by the friction force upon the system at time t i.e. m(t) per unit second is equal and opposite to that upon the sand particle per unit second. How to show it?

The sand particle contacts n points of the system at time t i.e. m(t) of length dr such that n dr = l the distance traveled by the particle before its speed becomes equal to v. I assume that this process takes 1s time.

The friction force acts upon each of the n points of the system at time t i.e. m(t) till each point move a distance dr in the opposite direction of the friction force.

Thus, the work done by the friction force upon the system at time t i.e. m(t) per unit second is $W_{cb} = f_{r_{cb}} n dr = - f_{r_{sp}} l = - \frac 1 2 1.5 v^2$ , where $f_{r_{cb} }$is the friction force acting upon the system at time t i.e. m(t) and $f_{r_{sp}}$ is the friction force acting upon the sand particle.

Now, (work done by me on the system at time t i.e. m(t) to keep it moving with speed v + work done by friction force on the system at time t i.e. m(t) ) per unit second = ( change in the kinetic energy of the system at time t i.e. m(t) per unit time=0).

change in the kinetic energy of the system at time t i.e. m(t) per unit time=0 as the speed of the system at time t i.e. m(t) remains same.

Required power = (work done by me on the system at time t i.e. m(t) to keep it moving with speed v) per unit second

= - ( work done by friction force on the system at time t i.e. m(t) ) per unit second = $\frac 1 2 \frac { dm} {dt} v^2 = 36.75$ watt.

 

[Pushoam]

Momentum change of the system due to friction will be zero owing to its internal nature but heat loss would have to be considered

Every sand particle doesn't start moving as soon as it falls...but over a short interval it has to overcome friction and attain the velocity...out of 73.5 watt 36.75 goes into kinetic energy and remaining half goes into work against friction

I want to arrive at what you are saying, mathematically, but getting stuck.

 

[jbriggs444]

I want to arrive at what you are saying, mathematically, but getting stuck.

Rather than chasing after an energy balance, try chasing the momentum balance. As has been pointed out, the problem with mechanical energy is that it is not conserved -- it can drain into frictional losses.

So... at what rate does momentum have to be delivered to the stream of sand?

Edit: Silly me, you've been down this road in the OP

 

[CWatters]

$P = \vec F \cdot \vec v = \frac{ dm} {dt} \vec v \cdot \vec v = \frac { dm} {dt} v^2 = 73.5$ watt...(2)

I see you replace F with dm/dt * v

The velocity isn't constant, it goes from 0 to v so the average is 0.5v. This looks to be where the factor of 2 error is coming in.