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Blazing goIITian

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6 Oct 2008 21:10:36 IST

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Kinematics

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Q. A particle of mass 10^-2kg is moving along the positive x axis under the influence of a force F(x)= -k/(2x²) where k=10^-2Nm². at time t=0 it is at x=1.0m and its velocity is v=0. Find

(a) its velocity when it reaches x=0.50m

(b) the time at which it reaches x=0.25m.

[Ans. (a) -1m/s (b) sec.]

Kindly , Explain each step.........Rates assured......

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Comments (6)

Priyesh

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6 Oct 2008 21:41:12 IST

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a) work done = change in kinetic enrgy

work done from x = 1m to x =0.5m = integral -k/2x^2dx (x from 1 to .5)

= [k/2 * 1/x] = k/2

=>k/2 = 1/2mv^2 => v^2 = k/m =.01/.01 = 1

since intitially particle is at rest & acceleration is towards left hence final velocity is also left hence v = - 1m/sec

Ragadeepika Pucha

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Joined: 12 Apr 2008

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7 Oct 2008 00:12:57 IST

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Given

Put F=ma -->

Solving this we get,

Note that the body is moving towards -ve X-axis..Hence velocity will also be negative.

v = -1m/s

I am sorry that I couldn't help you in part(b)

Hope this helps

ALL THE BEST

kushi

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Joined: 3 Aug 2008

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7 Oct 2008 03:49:36 IST

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reson given by above person for -ve velocity is not correct as it is not given in ques that particle is moving along -ve x axis

here velocity we choose is of -ve sign because velocity of particle is in opp dirn of force applied so acceleration also opp to velocity hence velocity of particle is decreasing and taken as -ve

BALGANESH

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7 Oct 2008 10:40:31 IST

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b ) F=-k/2x^2
ma=-k/2x^2
vdv/dx = -k/2mx^2
integreate both sides with th elimits 0-v and 1-.25

we get v=root3
now v=dx/dt
root3 dt=dx
integrate with limit 0-t for time and 1- 0.25 for x
we get t =root3/4 + c
c=pie/3

edison

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Joined: 19 Oct 2006

Posts: 7537

7 Oct 2008 12:05:41 IST

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Hope your query is well addressed and answered as above

vicor

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Joined: 21 Aug 2008

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7 Oct 2008 14:45:41 IST

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i hope thr is no need of my answer now

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