Question

A solid sphere of mass \(2\,kg\) rolls on a horizontal surface at \(10\,m/s\) and then rolls up a \(30^\circ\) incline. The maximum height reached is:

• A. \(10\,m\)
• B. \(4.9\,m\)
• C. \(14.2\,m\)
• D. \(7.1\,m\)

Hint:

Rolling body \(\Rightarrow\) include rotational KE.

Answer 1

The Correct Option is D

To determine the maximum height reached by the sphere on the incline, we consider the conservation of energy. Initially, the sphere has kinetic energy due to both its translational motion and its rotational motion.

Step-by-Step Solution:

1. Calculate the translational kinetic energy \( \left( KE_{\text{trans}} \right) \) of the sphere:
   • The translational kinetic energy is given by \(KE_{\text{trans}} = \frac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the velocity.
   • Substitute values: \(\frac{1}{2} \times 2\,\text{kg} \times (10\,\text{m/s})^2 = 100\,\text{J}\).
2. Calculate the rotational kinetic energy \( \left( KE_{\text{rot}} \right) \) of the sphere:
   • For a solid sphere, the moment of inertia \(I\) is \(\frac{2}{5}mr^2\).
   • The rotational kinetic energy is given by \(KE_{\text{rot}} = \frac{1}{2}I\omega^2\).
   • Angular velocity \(\omega\) is related to linear velocity by \(v = r\omega\), hence \(\omega = \frac{v}{r}\).
   • Substitute \(I\) and \(\omega\): \(KE_{\text{rot}} = \frac{1}{2} \times \frac{2}{5}mr^2 \times \left(\frac{v}{r}\right)^2 = \frac{1}{5}mv^2\).
   • Substitute values: \(\frac{1}{5} \times 2\,\text{kg} \times (10\,\text{m/s})^2 = 40\,\text{J}\).
3. Calculate the total initial kinetic energy:
   • \(KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = 100\,\text{J} + 40\,\text{J} = 140\,\text{J}\).
4. When the sphere reaches its maximum height on the incline, all of its kinetic energy has been converted into gravitational potential energy \( \left( PE \right) \):
   • Potential energy is given by \(PE = mgh\), where \(h\) is the height.
   • Setting \(PE = KE_{\text{total}}\): \(mgh = 140\,\text{J}\).
5. Solve for \(h\):
   • Rearrange the equation: \(h = \frac{140\,\text{J}}{mg} = \frac{140}{2 \times 9.8}\).
   • Calculate: \(h = \frac{140}{19.6} \approx 7.14\,\text{m}\).

The maximum height reached by the sphere on the incline is approximately \(7.1\,\text{m}\). Thus, the correct answer is 7.1 m.