Question

If a solid sphere of mass \(2\) kg is rolling without slipping on a surface with a velocity of \[ 10\ \mathrm{ms^{-1}}, \] then the total kinetic energy of the sphere is

• A. \(70\) J
• B. \(140\) J
• C. \(280\) J
• D. \(350\) J

Hint:

For a solid sphere rolling without slipping, \[ \boxed{ K=\frac12mv^2+\frac12I\omega^2 =\frac{7}{10}mv^2. } \] Always use \[ \boxed{\omega=\frac vr.} \]

Answer 1

The Correct Option is B

Step 1: Write the expression for total kinetic energy. For a body rolling without slipping, \[ K = \frac12mv^2 + \frac12I\omega^2. \] For a solid sphere, \[ I=\frac25mr^2, \] and \[ \omega=\frac vr. \]

Step 2: Calculate the rotational kinetic energy. Substituting, \[ \frac12I\omega^2 = \frac12\left(\frac25mr^2\right)\left(\frac{v^2}{r^2}\right) = \frac15mv^2. \] Hence, \[ K = \frac12mv^2+\frac15mv^2 = \frac{7}{10}mv^2. \]

Step 3: Find the total kinetic energy. Given, \[ m=2\text{ kg}, \qquad v=10\text{ ms}^{-1}. \] Therefore, \[ K = \frac{7}{10}\times2\times100 = 140\text{ J}. \] Hence, \[ \boxed{140\text{ J}}. \] Thus, \[ \boxed{(B)} \] is the correct answer.