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 × 10⁻⁴ J by the end of the second revolution after the beginning of the motion?(a) 0.1 m/s²(b) 0.15 m/s²(c) 0.18 m/s²(d) 0.2 m/s²

Answer 1

Answer:-

m = 10 g = 0.01 kg

K.E. = 1/2 mv^2 = 8 × 10⁻⁴  ......... (1)

Put m = 0.01 in (1)

v^2 = 16 x 10 ^ (-2)   ------- (2)

v^2 = u^2 + 2as --Newton's 3rd law

Initial velocity u = 0

So, v^2 = 2 as

     v^2 = 2 a (4 pi r)

    a = v^2 / 8 pi r

Substituting V^2 and r in above.

a = 0.1 m/s²

Answer 2

Answer:

kinetic energy = 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