1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinetic Energy, work and force! Sand conveyor problem

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data
    In the attachment!


    2. Relevant equations
    P=F*v=F*d/t=Work/t
    KE=(mv^2)/2
    W=f*s

    3. The attempt at a solution
    For the first question, I Found it rather difficult to figure out WHAT exactly accelerated the sand. I would simply say the force of the engine? But at the same time in order for the sand falling to gain horizontal velocity it would need friction to keep it in place on the conveyor belt. But in what direction does the force of the friction act exactly? I feel like it'll just fall and plonk itself there. The force just holding it unto the conveyor belt!

    For the second question I'm more befuddled. To calculate this force I thought that perhaps I could use F= rate of change of momentum to solve it. So if initial velocity is 0 then the change will be 60*2=120. But then I find that perhaps one could use KE formula as well? I thought that if all the force goes into changing the velocity of the sand then i could backpedal (work=KE)/distance, but that yields 60N! half the amount! :S

    Calculating the power seems simple enough using the Power formula. But i'd either end up with 240 or 120, using the velocity 2 m/s and the 'KE-derived' force or the 'momentum' derived force.

    It later also asks why the rate of change of KE isn't the same as the power? I peeked at the answer after thinking long and hard and they wrote that it's due to loss of energy due to friction, gain in internal energy of conveyer belt. What confuses me with this is that I thought that the KE formula doesn't take this loss into account, or does it? I thought the KE formula would calculate the work change assuming that ALL the energy goes to change the object's KE, nothing else!

    To round off. I'm having trouble resolving the ideas of KE and F*d=W. I thought that the KE of the sand rate of change would be equal to F*2/time and also to the Power. Can anybody explain this to me in a concise manner? Thankyou so much. This site's a lifesaver.
     

    Attached Files:

  2. jcsd
  3. Apr 12, 2012 #2

    tms

    User Avatar

    Without the friction, what would the sand do? That should tell you the direction of the frictional force.

    Don't just write down numbers, use symbols. That way you can see what is going on in your equations. Also, you must always indicate units; a number with no unit is utterly meaningless. If you do those things, you will see something missing in your numbers.
     
  4. Apr 12, 2012 #3

    tms

    User Avatar

    Reviewing my reply, perhaps it was not very clear. Start with the complete equation for force:
    [tex]\vec F = \frac{d\vec p}{dt}[/tex]
    and look at the system (belt plus sand) as a whole.
     
  5. Apr 13, 2012 #4
    Hi! You're right. I ought to write out my equations a little more properly. I've gone over the problem and think I'm getting to grips with understanding it. There's just one question that really still bugs me.

    If kinetic energy is the work done in giving an object a certain velocity, can I calculate the work done per second on the sand on the conveyor belt using the KE formula?

    So. mass* velocity^2/2= 60 kg (mass of sand) * 2 m/s *1/2 = 120 watts

    Alternatively, I would think you could also figure out the work done using the Power=force*velocity.
    Force being dp/dt= 60 kg *(2-0) m/s (change in velocity) / 1 second

    = 120 N.

    Then plugging that into P=F*v, Power= 120 N * 2 m/s= 240 W!

    My question then is why isn't the rate of change of kinetic energy equal to the power supplied? I understand that friction and gain in internal energy, etc. would cause dissipation of energy and therefore more power is required. But I thought both equations do not take these losses into account? Why is the KE rate of change any different from the power supplied?

    That's my great hangup! I'd be really thankful if you could enlighten me on that front? :) Thankyou!
     
  6. Apr 13, 2012 #5
    I always thought of KE as the work done in giving the object energy. So why can't one calculate the power using the kinetic energy formula/time, if gain in KE is nothing but work?
     
  7. Apr 13, 2012 #6
    you are correct to use force = rate of change of momentum to find the force.
    To calculate KE you need average force x velocity
    You are also correct to identify that the discrepancy in energy is due to work done against friction (heat)
    Correction: power = average force x velocity (KE per second)
     
    Last edited: Apr 13, 2012
  8. Apr 13, 2012 #7

    gneill

    User Avatar

    Staff: Mentor

    It might also help to think about the problem as a 'continuous' inelastic collision. You (should) know that kinetic energy is not conserved over inelastic collisions...
     
  9. Apr 15, 2012 #8

    tms

    User Avatar

    Again, look at the complete equation for force:
    [tex]\vec F = \frac{d\vec p}{dt}[/tex]
    Don't forget that, for the belt plus sand system, mass is not constant. That is,
    [tex]\vec F = \frac{d\vec p}{dt} = \vec v \frac{dm}{dt} + m \frac{d\vec v}{dt}[/tex]
    Look at the entire system, not just the sand falling onto the belt.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Kinetic Energy, work and force! Sand conveyor problem
Loading...