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Laws Of Motion

Question
CBSEENPH11020529

A particle of mass m is driven by a machine that delivers a constant power K watts. If the particle starts from rest, the force on the particle at time t is

  • square root of mk over 2 end root t to the power of negative 1 divided by 2 end exponent
  • square root of mk space t to the power of negative 1 divided by 2 end exponent
  • square root of 2 mk end root space t to the power of negative 1 divided by 2 end exponent
  • 1 half square root of m t end root to the power of negative 1 divided by 2 end exponent

Solution

A.

square root of mk over 2 end root t to the power of negative 1 divided by 2 end exponent

As the machine delivers a constant power
So F, v =constant = k (watts)
rightwards double arrow space straight m dv over dt. straight v space equals straight k
rightwards double arrow integral vdv space equals straight k over straight m integral dt
rightwards double arrow straight v squared over 2 equals straight k over straight m straight t space space
rightwards double arrow space straight v equals space square root of fraction numerator 2 straight k over denominator straight m end fraction end root straight t
Now comma space force space on space the space particles space is space given space by
straight F equals space straight m fraction numerator d straight v over denominator d straight t end fraction space equals straight m straight d over dt open parentheses fraction numerator 2 kt over denominator straight m end fraction close parentheses to the power of 1 divided by 2 end exponent
equals space square root of 2 km end root space open parentheses 1 half straight t to the power of negative 1 half end exponent close parentheses
equals space square root of mk over 2 end root. straight t to the power of fraction numerator negative 1 over denominator 2 end fraction end exponent