Question

A particle of mass $10g$ moves along a circle of radius $6.4cm$ 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. $0.1m/s^2$
• B. $0.15m/s^2$
• C. $0.18m/s^2$
• D. $0.2m/s^2$

Answer 1

Answer: (A)

$0.1m/s^2$

Energy, $E=\frac{1}{2}m{v}^2$

$\Rightarrow 8\times {10}^{-4}=\frac{1}{2}\left(\frac{10}{1000}\right)v^2$

$\Rightarrow v^2=16\times {10}^{-2}$

$\Rightarrow v=4\times {10}^{-1}=0.4m/s$

Now, $v^2=u^2+2a_{t}s$ where, $s=4\pi r$

$\Rightarrow \frac{16}{100}=0+2a_t(4\times \frac{22}{7}\times \frac{6.4}{100})$

$\Rightarrow a_t=0.1m/s^2$