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 $8\times {10}^{-4}$ J boy the end of the second revolution after the beginning of the motion?

• A. $0.2{m/s}^2$
• B. $0.1{m/s}^2$
• C. $0.15{m/s}^2$
• D. $0.18{m/s}^2$

Answer 1

The correct option is B $0.1{m/s}^2$

kinetic energy =$8\times {10}^{-4}$J

or, $\frac{1}{2}mv²=8\times {10}^{-4}$

or, $\frac{1}{2}\times 10\times {10}^{-3}v²=8\times {10}^{-4}$

or, $v²=16\times {10}^{-2}=>v=0.4m/s$

initial velocity of particle, $u=0m/s$

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

total distance covered, $s=2\left(2\pi r\right)=4\pi r$

so, $v^2=2as$

$a=\frac{v^2}{2s}=\frac{(0.4{)}^2}{2\left(4\pi r\right)}$

$=\frac{16\times {10}^{-2}}{(8\times 3.14\times 6.4\times {10}^{-}2)}$

$=0.0995m/s²\approx 0.1m/s²$