Question
What is the translational speed of a bowling ball at the top of a 0.76 m rise, given an initial translational speed of 5.40 m/s at the bottom?
Answers
Answered by
GPT-4o mini
To calculate the translational speed of a bowling ball at the top of a 0.76 m rise, we can use the principle of conservation of energy. Assuming no energy is lost to friction or air resistance, the total mechanical energy at the bottom of the rise (kinetic energy) will be equal to the total mechanical energy at the top of the rise (the sum of kinetic energy and potential energy).
1. **Initial Conditions:**
- The initial translational speed at the bottom (\(v_0\)) = 5.40 m/s
- The height of the rise (\(h\)) = 0.76 m
- Mass of the bowling ball (\(m\)) is not required since it will cancel out.
2. **Kinetic Energy at the Bottom:**
\[
KE_{initial} = \frac{1}{2} m v_0^2 = \frac{1}{2} m (5.40)^2
\]
3. **Potential Energy at the Top:**
\[
PE_{top} = mgh = mg(0.76)
\]
4. **Kinetic Energy at the Top:**
Let \(v\) be the speed at the top.
\[
KE_{top} = \frac{1}{2} m v^2
\]
5. **Conservation of Energy:**
According to the conservation of energy:
\[
KE_{initial} = PE_{top} + KE_{top}
\]
Substituting the expressions we have:
\[
\frac{1}{2} m (5.40)^2 = mg(0.76) + \frac{1}{2} m v^2
\]
6. **Canceling \(m\) from both sides (since it is non-zero):**
\[
\frac{1}{2} (5.40)^2 = g(0.76) + \frac{1}{2} v^2
\]
7. **Plug in \(g\):**
Using \(g \approx 9.81 \, \text{m/s}^2\):
\[
\frac{1}{2} (5.40)^2 = 9.81(0.76) + \frac{1}{2} v^2
\]
8. **Calculate \( \frac{1}{2} (5.40)^2 \):**
\[
\frac{1}{2} (5.40)^2 = \frac{1}{2} (29.16) = 14.58
\]
9. **Calculate \(9.81(0.76)\):**
\[
9.81 \times 0.76 = 7.4536
\]
10. **Substituting back:**
\[
14.58 = 7.4536 + \frac{1}{2} v^2
\]
11. **Solving for \(\frac{1}{2} v^2\):**
\[
14.58 - 7.4536 = \frac{1}{2} v^2
\]
\[
7.1264 = \frac{1}{2} v^2
\]
\[
v^2 = 2 \times 7.1264 = 14.2528
\]
12. **Calculate \(v\):**
\[
v = \sqrt{14.2528} \approx 3.78 \, \text{m/s}
\]
Thus, the translational speed of the bowling ball at the top of the 0.76 m rise is approximately **3.78 m/s**.
1. **Initial Conditions:**
- The initial translational speed at the bottom (\(v_0\)) = 5.40 m/s
- The height of the rise (\(h\)) = 0.76 m
- Mass of the bowling ball (\(m\)) is not required since it will cancel out.
2. **Kinetic Energy at the Bottom:**
\[
KE_{initial} = \frac{1}{2} m v_0^2 = \frac{1}{2} m (5.40)^2
\]
3. **Potential Energy at the Top:**
\[
PE_{top} = mgh = mg(0.76)
\]
4. **Kinetic Energy at the Top:**
Let \(v\) be the speed at the top.
\[
KE_{top} = \frac{1}{2} m v^2
\]
5. **Conservation of Energy:**
According to the conservation of energy:
\[
KE_{initial} = PE_{top} + KE_{top}
\]
Substituting the expressions we have:
\[
\frac{1}{2} m (5.40)^2 = mg(0.76) + \frac{1}{2} m v^2
\]
6. **Canceling \(m\) from both sides (since it is non-zero):**
\[
\frac{1}{2} (5.40)^2 = g(0.76) + \frac{1}{2} v^2
\]
7. **Plug in \(g\):**
Using \(g \approx 9.81 \, \text{m/s}^2\):
\[
\frac{1}{2} (5.40)^2 = 9.81(0.76) + \frac{1}{2} v^2
\]
8. **Calculate \( \frac{1}{2} (5.40)^2 \):**
\[
\frac{1}{2} (5.40)^2 = \frac{1}{2} (29.16) = 14.58
\]
9. **Calculate \(9.81(0.76)\):**
\[
9.81 \times 0.76 = 7.4536
\]
10. **Substituting back:**
\[
14.58 = 7.4536 + \frac{1}{2} v^2
\]
11. **Solving for \(\frac{1}{2} v^2\):**
\[
14.58 - 7.4536 = \frac{1}{2} v^2
\]
\[
7.1264 = \frac{1}{2} v^2
\]
\[
v^2 = 2 \times 7.1264 = 14.2528
\]
12. **Calculate \(v\):**
\[
v = \sqrt{14.2528} \approx 3.78 \, \text{m/s}
\]
Thus, the translational speed of the bowling ball at the top of the 0.76 m rise is approximately **3.78 m/s**.