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A PARTICLE OF MASS m IS MOVING ON A STRAIGHT LINE SUCH THAT ITS SPEED CHANGES WITH DISTANCE TRAVELLED AS v = b x^1/2 . THE WORK DONE BY NET FORCE ON THE PARTICLE DURING FIRST t seconds what ?

RAGINI SEIWAL , 7 Years ago
Grade 12
anser 1 Answers
venkat

Last Activity: 7 Years ago

Given,
Mass of the particle = m
Velocity varies linearly with distance travelled as
      v=bx1/2
Force acting on the body,
  F=ma
  F=m*dv/dt
  F=m*(dv/dx)*(dx/dt)  [Multiplying and dividing with ‘dx’]
  F=mv*(dv/dx)  [\becausedx/dt=v]
  F=mv*(d(bx1/2)/dx)
  F=m(bx1/2)(bx-1/2/2)
  F=mb2/2
  Distance moved by by the body in first ‘t’ seconds is given by
  v=dx/dt
  dt=dx/v
  Integrating on both sides, we get
\int_{}^{}dt=\intdx/v
  t=(1/b)\intdx/(x1/2)
  t=(2/b)*x1/2
  t=(2x1/2)/b
  i.e., distance travelled by the body in first t seconds  s=b2t2/4
  Work done by the force on the body in first t seconds is given by
  W=Fs
  W={mb2/2}*{b2t2/4}
 \therefore Work done W=mb4t2/8

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