Question

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?
[NEET-2016]​

Answer 1

See the above attachment.

Answer 2

Answer:

The kinetic energy is given as = 8 × 10-4 J

or, 1/2 mv² = 8 × 10-4

or, 1/2 × 10 × 10-3 v² = 8 × 10-4

or, v² = 16 × 10-2 => v = 0.4 m/s

initial velocity of particle, u = 0 m/s

we have to find Tangential acceleration at the end of 2nd revolution.

total distance covered, s = 2(2πr) = 4πr

so, v² = 2as

a = v²/2s = (0.4)²/2(4πr)

= 16 × 10^-2/(8 × 3.14 × 6.4 × 10-2)

= 0.0995 m/s² ≈ 0.1 m/s²

Explanation:

hope u get the answer btw thnks for marking my answer brainliest