The electric motor of a model train accelerates the train from rest to 0.760 m/s in 20.0 ms. The total mass of the train is 545 g. Find the average power delivered to the train during its acceleration.
Simple harmonic motion
Simple harmonic motion is a type of periodic motion in which an object undergoes oscillatory motion. The restoring force exerted by the object exhibiting SHM is proportional to the displacement from the equilibrium position. The force is directed towards the mean position. We see many examples of SHM around us, common ones are the motion of a pendulum, spring and vibration of strings in musical instruments, and so on.
Simple Pendulum
A simple pendulum comprises a heavy mass (called bob) attached to one end of the weightless and flexible string.
Oscillation
In Physics, oscillation means a repetitive motion that happens in a variation with respect to time. There is usually a central value, where the object would be at rest. Additionally, there are two or more positions between which the repetitive motion takes place. In mathematics, oscillations can also be described as vibrations. The most common examples of oscillation that is seen in daily lives include the alternating current (AC) or the motion of a moving pendulum.
![**Problem Statement:**
*The electric motor of a model train accelerates the train from rest to 0.760 m/s in 20.0 ms. The total mass of the train is 545 g. Find the average power delivered to the train during its acceleration.*
**Solution:**
Given:
- Initial velocity (\(u\)) = 0 m/s (since the train starts from rest)
- Final velocity (\(v\)) = 0.760 m/s
- Time (\(t\)) = 20.0 ms = 0.020 s (conversion from milliseconds to seconds)
- Mass (\(m\)) = 545 g = 0.545 kg (conversion from grams to kilograms)
1. **Calculate the acceleration (\(a\)) of the train:**
\[
a = \frac{v - u}{t} = \frac{0.760 \, \text{m/s} - 0 \, \text{m/s}}{0.020 \, \text{s}} = 38 \, \text{m/s}^2
\]
2. **Calculate the force (\(F\)) exerted on the train:**
\[
F = m \times a = 0.545 \, \text{kg} \times 38 \, \text{m/s}^2 = 20.71 \, \text{N}
\]
3. **Calculate the work done (\(W\)) on the train:**
Work done is the change in kinetic energy.
\[
W = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.545 \, \text{kg} \times (0.760 \, \text{m/s})^2 = 0.157 \, \text{J}
\]
4. **Calculate the average power (\(P\)) delivered to the train:**
\[
P = \frac{W}{t} = \frac{0.157 \, \text{J}}{0.020 \, \text{s}} = 7.85 \, \text{W}
\]
Therefore, the average power delivered to the train during its acceleration is **7.85 W**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa7224418-dab7-4e1b-89e5-af8c3fbf27a9%2Fb99e46be-d96e-403d-9560-fa3d7f39381f%2Fdk0wl28_processed.png&w=3840&q=75)
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