A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to 8 x 10-4 J by the end of the second revolution after the beginning of the motion? from Physics Motion in A Plane Class 11 Manipur Board
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A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to 8 x 10-4 J by the end of the second revolution after the beginning of the motion?

  • 0.15 m/s2

  • 0.18 m/s2

  • 0.2 m/s2

  • 0.1 m/s2


D.

0.1 m/s2

Given, mass of particle. m = 0.01 kg

Radius of circle along which particle is moving , r = 6.4 cm

Kinetic energy of particle, K.E. = 8 x 10-4 J

rightwards double arrow space 1 half mv squared space equals space 8 space straight x space 10 to the power of negative 4 end exponent space straight J

rightwards double arrow space straight v squared space equals space fraction numerator 16 space straight x space 10 to the power of negative 4 end exponent over denominator 0.01 end fraction
space space space space space space space space space space space equals 16 straight x 10 to the power of negative 2 end exponent space space space space space space space space space space space... space left parenthesis straight i right parenthesis

Given that, KE of particle is equal to 8 x 10-4 J by the end of second revolution after the beginning of the motion of particle.

It means, initial velocity (u) is 0 m/s at this moment.

Now, using the Newton's 3rd equation of motion,

v2 = u2 + 2as

rightwards double arrow space space space straight v squared space equals space 2 straight a subscript straight t straight s

rightwards double arrow space straight v squared space equals space 2 straight a subscript straight t space left parenthesis 4 πr right parenthesis

rightwards double arrow space straight a subscript straight t space equals space fraction numerator straight v squared over denominator 8 πr end fraction
space space space space space space space space space equals space fraction numerator 16 space straight x space 10 to the power of negative 2 end exponent over denominator 8 space straight x space 3.14 space straight x space 6.4 space straight x space 10 to the power of negative 2 end exponent end fraction

space space space space straight a subscript straight t space equals space 0.1 space straight m divided by straight s squared

Given, mass of particle. m = 0.01 kg

Radius of circle along which particle is moving , r = 6.4 cm

Kinetic energy of particle, K.E. = 8 x 10-4 J

rightwards double arrow space 1 half mv squared space equals space 8 space straight x space 10 to the power of negative 4 end exponent space straight J

rightwards double arrow space straight v squared space equals space fraction numerator 16 space straight x space 10 to the power of negative 4 end exponent over denominator 0.01 end fraction
space space space space space space space space space space space equals 16 straight x 10 to the power of negative 2 end exponent space space space space space space space space space space space... space left parenthesis straight i right parenthesis

Given that, KE of particle is equal to 8 x 10-4 J by the end of second revolution after the beginning of the motion of particle.

It means, initial velocity (u) is 0 m/s at this moment.

Now, using the Newton's 3rd equation of motion,

v2 = u2 + 2as

rightwards double arrow space space space straight v squared space equals space 2 straight a subscript straight t straight s

rightwards double arrow space straight v squared space equals space 2 straight a subscript straight t space left parenthesis 4 πr right parenthesis

rightwards double arrow space straight a subscript straight t space equals space fraction numerator straight v squared over denominator 8 πr end fraction
space space space space space space space space space equals space fraction numerator 16 space straight x space 10 to the power of negative 2 end exponent over denominator 8 space straight x space 3.14 space straight x space 6.4 space straight x space 10 to the power of negative 2 end exponent end fraction

space space space space straight a subscript straight t space equals space 0.1 space straight m divided by straight s squared

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