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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
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.
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