To find the translational kinetic energy of a disc rolling down an inclined plane, we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:
### Step 1: Identify the Given Information
- Mass of the disc, \( m = 3 \, \text{kg} \)
- Height of the inclined plane, \( h = 5 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
### Step 2: Calculate the Initial Potential Energy
The potential energy (PE) at the top of the inclined plane can be calculated using the formula:
\[
PE = mgh
\]
Substituting the values:
\[
PE = 3 \, \text{kg} \times 10 \, \text{m/s}^2 \times 5 \, \text{m} = 150 \, \text{J}
\]
### Step 3: Apply Conservation of Energy
At the top of the incline (point A), the disc has potential energy and no kinetic energy (since it starts from rest). At the bottom of the incline (point B), all the potential energy is converted into kinetic energy (KE). The total mechanical energy at both points is conserved:
\[
PE_A + KE_A = PE_B + KE_B
\]
Since \( KE_A = 0 \) and \( PE_B = 0 \), we have:
\[
PE_A = KE_B
\]
### Step 4: Express Kinetic Energy at the Bottom
The kinetic energy at the bottom consists of translational kinetic energy and rotational kinetic energy. The translational kinetic energy (TKE) is given by:
\[
TKE = \frac{1}{2} m V^2
\]
The rotational kinetic energy (RKE) for a disc is given by:
\[
RKE = \frac{1}{2} I \omega^2
\]
For a solid disc, the moment of inertia \( I \) about its center is:
\[
I = \frac{1}{2} m R^2
\]
And since the disc rolls without slipping, we have:
\[
\omega = \frac{V}{R}
\]
Substituting \( \omega \) into the RKE formula:
\[
RKE = \frac{1}{2} \left(\frac{1}{2} m R^2\right) \left(\frac{V^2}{R^2}\right) = \frac{1}{4} m V^2
\]
### Step 5: Set Up the Energy Equation
Now, we can express the total kinetic energy at the bottom:
\[
KE_B = TKE + RKE = \frac{1}{2} m V^2 + \frac{1}{4} m V^2 = \frac{3}{4} m V^2
\]
Setting this equal to the potential energy at the top:
\[
150 \, \text{J} = \frac{3}{4} m V^2
\]
### Step 6: Solve for \( V^2 \)
Substituting \( m = 3 \, \text{kg} \):
\[
150 = \frac{3}{4} \times 3 V^2
\]
\[
150 = \frac{9}{4} V^2
\]
Multiplying both sides by \( \frac{4}{9} \):
\[
V^2 = \frac{150 \times 4}{9} = \frac{600}{9} \approx 66.67
\]
### Step 7: Calculate Translational Kinetic Energy
Now substituting \( V^2 \) back into the translational kinetic energy formula:
\[
TKE = \frac{1}{2} m V^2 = \frac{1}{2} \times 3 \times \frac{600}{9}
\]
\[
TKE = \frac{3 \times 600}{18} = \frac{1800}{18} = 100 \, \text{J}
\]
### Final Answer
The translational kinetic energy of the disc on reaching the bottom of the inclined plane is:
\[
\boxed{100 \, \text{J}}
\]
