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a particle of mass m starts moving from origin along x axis and its velocity varies with position (x) as v=k√x. the work done by force acting on it during first "t" seconds is ???ans) (m k^4 t^2) ÷ 8

a particle of mass m starts moving from origin along x axis and its velocity varies with position (x) as v=k√x. the work done by force acting on it during first "t" seconds is ???ans) (m k^4 t^2) ÷ 8

Grade:11

7 Answers

DR STRANGE
212 Points
7 years ago
HI Vaibhav,
here you have to apply some calculus,
v=\frac{dx}{dt}=k\sqrt{x},
acceleration,
a =\frac{d(v)}{dt}=\frac{d(k\sqrt{x})}{dt}=\frac{k}{2\sqrt{x}}\frac{dx}{dt}=\frac{k}{2\sqrt{x}}v =\frac{k}{2\sqrt{x}}\frac{k\sqrt{x}}{2}=\frac{k^{2}}{4}
distance travelled in time t is ,
dx=k\sqrt{x} dt
\frac{dx}{\sqrt{x}}=k dt
\int \frac{dx}{\sqrt{x}}=\int k dt
2\sqrt{x}=kt    
x=\frac{k^{2}t^{2}}{4}
workdone=F\times s=m\times a\times s
=m\times \frac{k^{2}}{2}\times \frac{k^{2}t^{2}}{4}=\frac{mk^{4}t^{2}}{8}
…...............................ANS
APPROVE IF YOU ARE SATISFIED :P
 
Vaibhav
35 Points
7 years ago
I want you to recheck your answer mr strange...you had written the value of a as k^2 ÷ 4 and you put that value at last as k^2 ÷ 2....Overall if u cann xplain it in brief i wud be thankful.... i also dont understan how did this came [k√x÷2]
DR STRANGE
212 Points
7 years ago
sry it was a mistake ,
a =\frac{d(v)}{dt}=\frac{d(k\sqrt{x})}{dt}=\frac{k}{2\sqrt{x}}\frac{dx}{dt}=\frac{k}{2\sqrt{x}}v =\frac{k}{2\sqrt{x}}\frac{k\sqrt{x}}{1}=\frac{k^{2}}{2}
this is right.......... :)
…...........................................................................................
 
Vaibhav
35 Points
7 years ago
I dont understand how did you got (k dx)/(2√x dt).....and also this the integration of dx/√x should be x/2 but u wrote 2x how???
Ashmak Moon
27 Points
7 years ago
Here's some simpler method you can use:
 (Assuming K is a constant and indepenfent of time);a =\frac{d(v)}{dt}=\frac{d(k\sqrt{x})}{dt}=\frac{k}{2\sqrt{x}}\frac{dx}{dt}=\frac{k}{2\sqrt{x}}v =\frac{k}{2\sqrt{x}}\frac{k\sqrt{x}}{2}=\frac{k^{2}}{4}
then we use 2nd equation of motion to get displacement as s = ut + 1/2 at^2 to egt displacement. Putting initial position and hence the velocity to be zero, we get s= (1/2)*(k2/4)*tand hence work done= m*a*s= m* (k2/4)* (1/2)*(k2/4)*t2=mk4t2/8
Ashmak Moon
27 Points
7 years ago
That was copied from above vaibhav, sorry for that, just put the correct method for á’ and you shall find the correct answer.
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.
645-322_02.PNG

Thanks and Regards

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