A uniform thin ring of mass 0.4 kg rolls without slipping on a horizontal surface with a linear velocity of 10 cm/s. The kinetic energy of the ring is

Question

A uniform thin ring of mass 0.4 kg rolls without slipping on a horizontal surface with a linear velocity of 10 cm/s. The kinetic energy of the ring is (A)  4× 10-3  (B)  4× 10-2J  (C)  2× 10-3J  (D)  2× 10-2J

Tardigrade Admin · Accepted Answer

Given, m=0.4 kg and v=10 cm/s Kinetic energy =(1/2)mv2(1+1) Kinetic energy =mv2 =0.4× (0.1)2 =4× 10-3J

Q. A uniform thin ring of mass 0.4 kg rolls without slipping on a horizontal surface with a linear velocity of 10 cm/s. The kinetic energy of the ring is

$4 \times 10^{- 3}$

$4 \times 10^{- 2} J$

$2 \times 10^{- 3} J$

$2 \times 10^{- 2} J$

Given, $m = 0.4 k g$ and $v = 10 c m / s$ Kinetic energy $= \frac{1}{2} m v^{2} (1 + 1)$ Kinetic energy $= m v^{2}$ $= 0.4 \times (0.1)^{2}$ $= 4 \times 10^{- 3} J$

Q. A uniform thin ring of mass 0.4 kg rolls without slipping on a horizontal surface with a linear velocity of 10 cm/s. The kinetic energy of the ring is

4×10−3

4×10−2J

2×10−3J

2×10−2J

Given, m=0.4kg and v=10cm/s Kinetic energy =21​mv2(1+1) Kinetic energy =mv2 =0.4×(0.1)2 =4×10−3J

$4 \times 10^{- 3}$

$4 \times 10^{- 2} J$

$2 \times 10^{- 3} J$

$2 \times 10^{- 2} J$

Given, $m = 0.4 k g$ and $v = 10 c m / s$ Kinetic energy $= \frac{1}{2} m v^{2} (1 + 1)$ Kinetic energy $= m v^{2}$ $= 0.4 \times (0.1)^{2}$ $= 4 \times 10^{- 3} J$