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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
one year ago

Answers : (6)

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
 
one year 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]
						
one year 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.......... :)
…...........................................................................................
 
one year 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???
						
one year 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)*tand hence work done= m*a*s= m* (k2/4)* (1/2)*(k2/4)*t2=mk4t2/8
one year 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.
one year ago
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