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Junior Member
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Jun 13, 2012, 09:59 AM
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Gain in kinetic energy calculation
Please help... a body of mass 5 kg is moving with a momentum of 10 kg m/s. a force of 0.2N acts on in the direction of motion of the body for 10 secs. The increase in K.E is..
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Expert
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Jun 13, 2012, 12:37 PM
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You can approach this in either of two ways:
1. Use , You can get started by determining the object's initial velocity, its acceleration, and then it's final velocity.
2. Use . To find d you will need to first determine the initial velocity and acceleration.
I suggest you try both approaches and see if you get the same answer. If you get stuck post back showing what you've tried and we'll help you out.
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Junior Member
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Jun 14, 2012, 12:43 AM
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I'm doing in this way...
K.E=F.S
=FVT
WHERE V FROM MOMENTUM WHICH IS 2 M/S
4 JOULE... BT ANS HERE IS 4.4 J...
another thing that increase in k.E is calculated from its simple formula(0.5 m v^2 or f.s) or shall I employ law of conservation of energy.. please guide..
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Expert
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Jun 14, 2012, 05:52 AM
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Originally Posted by subhan2010
]
K.E=F.S
=FVT
No - the distance the body travels wold equal vt only if v is constant, but it's not. Since a force is acting on it, there is acceleration (from F=ma), so its velocity changes over time. You need to use equations of motion to determine either the body's final velocity (if using method 1 above) or the distance traveled in 10 seconds (if using method 2):
In either case you must first determine the body's acceleration, from F=ma.
Originally Posted by subhan2010
WHERE V FROM MOMENTUM WHICH IS 2 M/S
Yes, v_1 = 2m/s.
Originally Posted by subhan2010
another thing that increase in k.E is calculated from its simple formula(0.5 m v^2 or f.s) or shall i employ law of conservation of energy...? plz guide..
Increase in KE can be found either from 1/2m(v_2^2 - v_1^2), or from the principal of work applied = change in energy, which is where Fs = change in KE comes from. These were the two methods I outlined earlier.
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Junior Member
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Jun 14, 2012, 06:49 AM
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Same ans is coming from your method.. 4 j... but in my book ans is 4.4 j..!
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Expert
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Jun 14, 2012, 06:55 AM
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Please show us the steps you are using to calculate acceleration 'a' , final velocity 'v_2', and change in KE.
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Junior Member
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Jun 14, 2012, 08:49 PM
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I am unable to calculate distance from EQUATIONS OF MOTION... plz show its working...
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Expert
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Jun 15, 2012, 05:38 AM
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Originally Posted by subhan2010
i am unable to calculate distance from EQUATIONS OF MOTION.......
Calculate distance traveled while the force is applied from:
You already know the values for and . You can get the value for from .
Post back with what you get for , and also the work done by the force: .
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Junior Member
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Jun 15, 2012, 06:13 AM
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I've got 4.4j... thankssss... last thing that how can u say that velocity calc. from momentum is initial.. can it be final.. explain..
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Junior Member
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Jun 15, 2012, 06:16 AM
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*cant
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Expert
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Jun 15, 2012, 06:46 AM
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Originally Posted by subhan2010
how can u say that velocity calc. from momentum is initial. .can it be final..explain...?
I don't understand what you're asking. Please use complete sentences so I can understand you. But if you're asking can the final velocity equal the initial velocity the answer is obvious - no. If you want to find the final velocity use
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Junior Member
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Jun 15, 2012, 09:48 AM
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no I'm nt asking that.. I m saying that I've calculated velocity from momentum... i.e l=mass*velocity. The velocity obtainted from this is initial velocity... can't it be final velocity..
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Expert
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Jun 15, 2012, 10:03 AM
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Originally Posted by subhan2010
can't it be final velocity...?
Sorry - I don't understand your question. You say you know that final velocity is not the same as initial velocity, then you ask if it can be the same. So please clarify your question.
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Junior Member
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Jun 16, 2012, 08:40 AM
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Can we obtain final velocity from momentum..
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Expert
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Jun 16, 2012, 12:01 PM
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Originally Posted by subhan2010
can we obtain final velocity from momentum...?
If you know what the final momentum is, sure. But how are you going to determine what the final momentum is?
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Junior Member
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Jun 16, 2012, 10:49 PM
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No I was not asking about this with respect to this problem...
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