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?
Answer & explanation
Correct answer: option 1
KE : \(\frac{1}{2}mv^2 = 8 × 10^{-4}\)
m = 10 x 10-3 kg
v2= 16 x 10-12
⇒ v = 0.4 m/s
Eq. of motion : \(v^2 = u^2 + 2aS\)
S = 6.4 x 10-2 m ; u = 0
∴ 0.4 × 0.4 = 2 × a × 2\(\pi\) × 6.4 ×10-2 × 2
⇒ a = 0.1 m/s2