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Physics Question on work, energy and power

A particle of mass 10 g10\,g moves along a circle of radius 6.4 cm6.4\,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 J8 \times 10^{-4} \, J by the end of the second revolution after the beginning of the motion ?

A

0.15 m/s20.15 \, m / s^2

B

0.18 m/s20.18 \, m / s^2

C

0.2 m/s20.2 \, m / s^2

D

0.1 m/s20.1 \, m / s^2

Answer

0.1 m/s20.1 \, m / s^2

Explanation

Solution

The correct option is(D): 0.1 m/s2
12mv2=8×10−4\frac{1}{2} mv ^{2}=8 \times 10^{-4}
or, 12×10×10−3v2=8×10−4\frac{1}{2} \times 10 \times 10^{-3} v ^{2}=8 \times 10^{-4}
or, v2=16×10−2⇒v=0.4m/sv ^{2}=16 \times 10^{-2} \Rightarrow v =0.4 m / s
initial velocity of particle, u=0m/su =0 m / s
we have to find Tangential acceleration at the end of 2 nd revolution.
total distance covered, s=2(2πr)=4πrs =2(2 \pi r )=4 \pi r
so, v2=2asv ^{2}=2 as
a=v22s=(0.4)22(4πr)a=\frac{v^{2}}{2 s}=\frac{(0.4)^{2}}{2(4 \pi r)}
=16×10−2(8×3.14×6.4×10−2)=\frac{16 \times 10^{-2}}{\left(8 \times 3.14 \times 6.4 \times 10^{-2}\right)}
=0.0995m/s2≈0.1m/s2=0.0995 m / s ^{2} \approx 0.1 m / s ^{2}