IIT JEE , AIEEE Forum - Physics , Mechanics

vvshinde_wda

Force displacement graph

anchitsaini

edison

This is a problem of variable force acting on the body

Now at x = 2m/s, v= 3m/s

Kinetic energy at intial point = 0

Thus, change in kinetic energy at x=2/s is = (1/2)mv²

or Change in KE = 0.5*2*9 = 9 J

Here work done on the body is converted into its kinetic energy

Thus work done = F.d

But force here is a function of displacement x as follows

dF/dx = (9-0)/(2-0) from theabove condition

or dF/dx = 4.5 N/m

Thus force at any position x is given by

Thus F(x) = 4.5 x Newton

Thus work done by the force on the body is given by

W = _{[ 0]}

^{[ 10]} F(x).dx = _{[ 0]}^{[ 10]} 4.5 x dx = 4.5 * 10²/2 = 225 J

So kinetic energy gained by the object = 225J

or 1/2 m v² = 225

or v² = 225 or v = 15m/s

Thus final velocity of the at x = 10m is 15m/s

anchitsaini

sir i couldn't get it !!
whats wrong in my method?

anchitsaini

karthik2007

@edison - Sir, I think you have made a mistake somewhere in your solution.

abrambenny

Pls Correct Me if I'm wrng...

The area under the force displacemnt graph gives the work done: Here it is 1/2 *10*6=30J. We know that the K.E. of the body at x=2 is 9J. Then the increase in k.e. frm that point onwards must be 30-9=21J.

Therefore 1/2 mv^2=21. On solving we get v=(21)^1/2=4.58 m s^-1

priyesh

from the graph F = (3/5)x

so work done from x = 2 to x = 8 is

_{[ 2]}

^{[8 ]}Fdx = 3/5_[2]^[8]xdx = 3/10[ 64 - 4] = 18 J

now this work done = change in kinetic energy = 1/2 * 2[v^2 - 3^2]

=> v^2 = 18 + 9 => v = root(27) = 5.2 m/sec

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