A solid sphere of diameter 0.1 m and 5 kg is rolling down an inclined plane with a speed of 4 m/s. The total kinetic energy of the sphere is :

Options

A. 28 J
B. 56 J
C. 84 J
D. 112 J

Answer

To find the total kinetic energy of a solid sphere rolling down an inclined plane, we need to consider both its translational and rotational kinetic energy.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Diameter of the sphere, \( d = 0.1 \, \text{m} \)
- Radius of the sphere, \( r = \frac{d}{2} = \frac{0.1}{2} = 0.05 \, \text{m} \)
- Mass of the sphere, \( m = 5 \, \text{kg} \)
- Speed of the sphere, \( v = 4 \, \text{m/s} \)

2. **Calculate the Translational Kinetic Energy (TKE):**
The formula for translational kinetic energy is:
\[ \text{TKE} = \frac{1}{2} mv^2 \]
Substituting the values:
\[ \text{TKE} = \frac{1}{2} \times 5 \times (4)^2 = \frac{1}{2} \times 5 \times 16 = \frac{5 \times 16}{2} = 40 \, \text{J} \]

3. **Calculate the Moment of Inertia (I) for a Solid Sphere:**
The moment of inertia for a solid sphere is given by:
\[ I = \frac{2}{5} m r^2 \]
Substituting the values:
\[ I = \frac{2}{5} \times 5 \times (0.05)^2 = \frac{2}{5} \times 5 \times 0.0025 = \frac{2 \times 5 \times 0.0025}{5} = 0.01 \, \text{kg m}^2 \]

4. **Calculate the Angular Velocity (ω):**
The angular velocity \( \omega \) can be related to the linear velocity \( v \) by:
\[ \omega = \frac{v}{r} \]
Substituting the values:
\[ \omega = \frac{4}{0.05} = 80 \, \text{rad/s} \]

5. **Calculate the Rotational Kinetic Energy (RKE):**
The formula for rotational kinetic energy is:
\[ \text{RKE} = \frac{1}{2} I \omega^2 \]
Substituting the values:
\[ \text{RKE} = \frac{1}{2} \times 0.01 \times (80)^2 = \frac{1}{2} \times 0.01 \times 6400 = 32 \, \text{J} \]

6. **Calculate the Total Kinetic Energy (TKE + RKE):**
\[ \text{Total Kinetic Energy} = \text{TKE} + \text{RKE} = 40 + 32 = 72 \, \text{J} \]

### Final Answer:
The total kinetic energy of the sphere is **72 joules**.