To determine the speed of the pendulum at the lowest point of its path, we will use the principles of conservation of energy. Specifically, we'll use the fact that the potential energy at the starting height is fully converted to kinetic energy at the lowest point.
Here’s the step-by-step solution:
1. Identify the Given Data:
- Mass of the pendulum, [tex]\( m = 1.5 \, \text{kg} \)[/tex]
- Initial height, [tex]\( h = 0.4 \, \text{m} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
2. Calculate the Potential Energy at the Starting Height:
Potential energy (PE) is given by the formula:
[tex]\[
\text{PE}_{\text{initial}} = mgh
\][/tex]
Substituting the values:
[tex]\[
\text{PE}_{\text{initial}} = 1.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.4 \, \text{m}
\][/tex]
[tex]\[
\text{PE}_{\text{initial}} = 5.880 \, \text{J}
\][/tex]
3. Relate Potential Energy to Kinetic Energy at the Lowest Point:
At the lowest point, all the initial potential energy is converted into kinetic energy (KE). Kinetic energy is given by the formula:
[tex]\[
\text{KE} = \frac{1}{2} mv^2
\][/tex]
Setting [tex]\(\text{PE}_{\text{initial}} = \text{KE}\)[/tex], we have:
[tex]\[
5.880 \, \text{J} = \frac{1}{2} \times 1.5 \, \text{kg} \times v^2
\][/tex]
4. Solve for the Velocity:
[tex]\[
5.880 = \frac{1}{2} \times 1.5 \times v^2
\][/tex]
Simplify the equation:
[tex]\[
5.880 = 0.75 \times v^2
\][/tex]
Solve for [tex]\( v^2 \)[/tex]:
[tex]\[
v^2 = \frac{5.880}{0.75}
\][/tex]
[tex]\[
v^2 = 7.840
\][/tex]
Taking the square root of both sides to find [tex]\( v \)[/tex]:
[tex]\[
v = \sqrt{7.840}
\][/tex]
[tex]\[
v \approx 2.8 \, \text{m/s}
\][/tex]
5. Conclusion:
The speed of the pendulum at the lowest point of its path is [tex]\( 2.8 \, \text{m/s} \)[/tex].
Therefore, the correct answer is:
A. [tex]\(2.8 \, \text{m/s}\)[/tex]
