 # under the action of a force ,a 2 kg body moves such that its position x as a function of time is given by x=t³/3.where t is in seconds and x in metre .the work done by force in first 2 seconds is? i diffrentiated x=t³/3 twice to get the equation for acceleration and got it 4 m/s^2 then force=4*2=8 N now as W=F.s i put W=8*t³/3 and tried integerating it both sides with limit 0 to 2 i can be horribly wrong but please tell me what i am doing wrong. thanks.

9 years ago

its very simpe bro

diff it you will get t^2

put t=2 in it you will get  4 thats v

the use Ke=1/2mv^2

=1/2x2x4x4
=16

9 years ago

differentiate it once to get velocityat t=o and t=2sec, uget u=0and v=4 m/s .apply we theorem to get work done =4N

sorry, calculation mistake answer would become 16N

9 years ago

As x is a function of time we can get the distance the body has travelled in the first two seconds. The one which u did to find acceleration of the body is right. As the force is 8N the displacement of the body is 8/3 so the work done is 64/3J. Or u can try it in another method. As u got the acceleration in terms of t which is 2t so F(t) is 4*2t which is 8t. Work by definition means F.dx. Find dx which we get it as t^2 dt now multiply F(t) with t^2 dt and integrate it by putting limits from 0 to 2 and now u will get the right answer which is 64/3J. Second way is a professional way of doing problems.

Hope this helps u.

4 years ago
Given,X= t^3/3.M= 2kgTime= 2 seconds.Differentiating,X=t^3/3, we get,X=t^2.Putting t=2 in K.E.= 1/2 MV^2, we getK.E= 1/2 X 2X4X4 = 16 NThe answer should be 16 joules
3 years ago
We know W=∆K
Differentiating x=t2/3 we get velocity
V =2t/3 putting t= 2 we get v=4/3
Now KE = 1/2mv2 =1/2•2•4/3•4/3 = 16/3
one year ago
Dear Student,

v = dx/dt​ = t^2
W = 0.5​mv^2
= 0.5​mt^4
= 0.5​×2×(2)^4
= 16J

Thanks and Regards