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Grade: 11 

                        
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
4 years ago

Answers : (7)

DR STRANGE
212 Points 
							
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
 
4 years ago
Vaibhav
35 Points 
							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]
4 years ago
DR STRANGE
212 Points 
							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.......... :)
…...........................................................................................
 
4 years ago
Vaibhav
35 Points 
							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???
4 years ago
Ashmak Moon
27 Points 
							
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)*t2 and hence work done= m*a*s= m* (k2/4)* (1/2)*(k2/4)*t2=mk4t2/8
4 years ago
Ashmak Moon
27 Points 
							
That was copied from above vaibhav, sorry for that, just put the correct method for á’ and you shall find the correct answer.
4 years ago
Rishi Sharma
askIITians Faculty
646 Points 
							Dear Student,
Please find below the solution to your problem.
645-322_02.PNG

Thanks and Regards
7 months ago
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