Question

A particle of a 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 by the end of the second revolution after the beginning of the motion?

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

Answer 1

The correct option is A 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, $v²=2as$

$a=\frac{v²}{2s}=\frac{(0.4{)}^2}{2\left(4\pi r\right)}$

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

$=0.0995m/s²\approx 0.1m/s²$