Grade 11General Physics

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

Vaibhav

9 Years agoGrade 11

7 Answers

DR STRANGE

9 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

9 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

9 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

9 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

9 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 2^nd 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)*t²and hence work done= m*a*s= m* (k²/4)* (1/2)*(k²/4)*t²=mk⁴t²/8