The Naked Scientists

The Naked Scientists Forum

Author Topic: calculate the instantaneous velocity  (Read 2570 times)

Offline Chikis

  • Jr. Member
  • **
  • Posts: 35
    • View Profile
calculate the instantaneous velocity
« on: 16/04/2014 06:32:36 »
A stone of mass 5 g is projected with a rubber catapult. If the catapult is streched through a distance of 7 cm by an average force of 70 N, calculate the instantaneous velocity of the stone when released.


mass m = 5g = 0.005kg; extension e = 7cm = 0.07m;
force f = 70 N; velocity = ?;
using:
1/2 fe = 1/2 mv2
v = square root ( fe/m)
v = (70 * 0.07/0.005)
= 31. 304 m/s
Using either of the equations gives the same result. Am not sure wether I have arrived because the book am using gave a different solution for the problem. So you watching at this thread, what do you think?


 

Offline daveshorts

  • Moderator
  • Neilep Level Member
  • *****
  • Posts: 2583
  • Physics, Experiments
    • View Profile
    • http://www.chaosscience.org.uk
Re: calculate the instantaneous velocity
« Reply #1 on: 16/04/2014 09:21:04 »
The energy in a spring which follows Hooke's law is 374f2b8a37400e1fa775155c0c2ae089.gif where f is the force at maximum extension, the 93b05c90d14a117ba52da1d743a43ab1.gif is essentially averaging the force.

You have been given the average force, so the energy in the spring is going to be 2d917f5d1275e96fd75e6352e26b1387.gif

Does that produce the speed you were expecting?
 

Offline Ocean Balodiya

  • First timers
  • *
  • Posts: 4
    • View Profile
Re: calculate the instantaneous velocity
« Reply #2 on: 16/04/2014 09:42:08 »
i think you have used the wrong formula...
basically what you should think is... that the work done to the catapult... will convert... totally into kinetic energy of the stone...
force in the formula (F e ) is F ..and  you are explicitly writing it 1/2 F... is already given to us as the avg...force.. we don't need to take the avg..of the avg force...
so...   F e = 1/2mv^2
and now.. you will get it right...  :)
 

Offline Chikis

  • Jr. Member
  • **
  • Posts: 35
    • View Profile
Re: calculate the instantaneous velocity
« Reply #3 on: 16/04/2014 22:17:54 »
i think you have used the wrong formula...
basically what you should think is... that the work done to the catapult... will convert... totally into kinetic energy of the stone...
force in the formula (F e ) is F ..and  you are explicitly writing it 1/2 F... is already given to us as the avg...force.. we don't need to take the avg..of the avg force...
so...   F e = 1/2mv^2
and now.. you will get it right...  :)

But where on earth did the expression 2d917f5d1275e96fd75e6352e26b1387.gif came from? What does fe mean here?
1cc1b53f02140ccf317b113177019dd8.gif

 = 44. 27 m/s
« Last Edit: 20/07/2014 15:17:34 by Chikis »
 

Offline Chikis

  • Jr. Member
  • **
  • Posts: 35
    • View Profile
Re: calculate the instantaneous velocity
« Reply #4 on: 16/04/2014 22:28:23 »
But if fe is work done, or elastic potential energy  of the catapult. Can we say 1/2fe = average work done
 

Offline Ocean Balodiya

  • First timers
  • *
  • Posts: 4
    • View Profile
Re: calculate the instantaneous velocity
« Reply #5 on: 17/04/2014 09:01:58 »
no... totally not. 1/2 F.e is  not avg.. work...
what would be the meaning of the AVG..work done...
say.. you move a stone placed on road... to a distance... x
and the force you applied is... F which is constant...
so.. the work done by you is...W= F.x
avg... work would mean nothing...
BUT... if say the force is variable... (changing with time)
and again you move the stone to distance.. x..
then the work W = {integration of.. F.x }over the proper interval...
in this case.. if you wanted a crude answer.. the you could average out the force that was acting on the stone...
to... F/2 .. and you could have got a very.. ... sort of unreliable answer
and in your question.. you are given ... average... force... that was being acted upon the stone...
 :)
 

The Naked Scientists Forum

Re: calculate the instantaneous velocity
« Reply #5 on: 17/04/2014 09:01:58 »

 

SMF 2.0.10 | SMF © 2015, Simple Machines
SMFAds for Free Forums
 
Login
Login with username, password and session length