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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
HI Vaibhav,here you have to apply some calculus,,acceleration,distance travelled in time t is , …...............................ANSAPPROVE IF YOU ARE SATISFIED :P
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]
sry it was a mistake ,this is right.......... :)…...........................................................................................
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???
Here's some simpler method you can use: (Assuming K is a constant and indepenfent of time);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
That was copied from above vaibhav, sorry for that, just put the correct method for á’ and you shall find the correct answer.
Dear Student,Please find below the solution to your problem.Thanks and Regards
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