A particle is projected with velocity V along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin. i.e, ma = -x 2 . The distance at which the particle stops
Hello candidate,
In order to solve this problem you have to take the help of differential form of acceleration in the form of velocity-
We know acceleration of the particle is equal to- a= -2x/m, and the formula acceleration is equal to vdv/dx=-2x/m, so the value of v equals to x/√m.
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Hi there,
F = –x2 ------(1)
ma = –x2-----(2)
a=-x2/m-----(3)
Substitute a=vdv/dx in the above equation (3), the equation becomes:
vdv/dx=-x2/m----(4)
Now integrate equation (4), to find x. The limits of integration will be between Vo and Zero. (initial and final respectively)
On solving you should get the value of x (distance) as (3Vo^2.m/2)^1/3
x= (3Vo^2.m/2)^1/3
I hope this helps!
Thank you!
