# calculate the instantaneous velocity

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#### Chikis

• Jr. Member
• 35
##### 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?

#### daveshorts

• Moderator
• Neilep Level Member
• 2583
• Physics, Experiments
##### 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 $$\frac{1}{2}fe$$ where f is the force at maximum extension, the $$\frac{1}{2}$$ is essentially averaging the force.

You have been given the average force, so the energy in the spring is going to be $$fe$$

Does that produce the speed you were expecting?

#### Ocean Balodiya

• First timers
• 4
##### 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...

#### Chikis

• Jr. Member
• 35
##### 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 $$fe$$ came from? What does fe mean here?
$$\sqrt{\frac{2\times70\times0.07}{0.005}}$$

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

#### Chikis

• Jr. Member
• 35
##### 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

#### Ocean Balodiya

• First timers
• 4
##### 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...
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