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 8×10-4 J by the end of the second revolution after the beginning of the motion?

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

Solution:

$Here, m = 10 g = 10^{- 2} kg$

$R = 6.4 cm = 6.4 \times 10^{- 2} m, K_{f} = 8 \times 10^{- 4} J, K_{i} = 0, a_{t} = ?$

$U \sin g w o r k - e n e r g y t h e o r e m, W o r k d o n e b y a l l t h e f o r c e s = C h a n g e i n K E$

$W_{tangential force} + W_{centripetal force} = K_{f} - K_{i}$

$\begin{matrix} \Rightarrow F_{t} \times s + 0 = K_{f} - 0 \Rightarrow m a_{t} \times (2 \times 2 π R) = K_{f} \\ \Rightarrow a_{t} = \frac{K_{f}}{4 π R m} = \frac{8 \times 10^{- 4}}{4 \times \frac{22}{7} \times 6.4 \times 10^{- 2} \times 10^{- 2}} \\ = 0.099 \approx 0.1 m s^{- 2} \end{matrix}$

A particle of mass $10 g$ moves along a circle of radius $6.4 c m$ 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?

Solution:

Related content

Company

Support

Courses

More

JEE

NEET

CUET

CBSE

Popular books

NCERT solutions

NCERT exemplar

State Board

Subject

follow us