What is the translational kinetic energy with a mass of 2000 [kg] travelling at the speed of 55 miles per hour?

Question 1 options:

3.1 x 10^5 [J]


3.025 x 10^6 [J]


1.25 x 10^6 [J]


6.25 x 10^5 [J]

To calculate the **translational kinetic energy (K.E.)**, we use the following formula:

\[
K.E. = \frac{1}{2} m v^2
\]

Where:
- \( m \) is the mass of the object in kilograms (kg),
- \( v \) is the velocity in meters per second (m/s).

### Step 1: Convert speed from miles per hour (mph) to meters per second (m/s).
- 1 mile = 1609.34 meters
- 1 hour = 3600 seconds

So,
\[
55 \, \text{mph} = 55 \times \frac{1609.34}{3600} \, \text{m/s} \approx 24.587 \, \text{m/s}
\]

### Step 2: Use the K.E. formula.
Now, substitute the mass \( m = 2000 \, \text{kg} \) and velocity \( v = 24.587 \, \text{m/s} \) into the kinetic energy formula:

\[
K.E. = \frac{1}{2} \times 2000 \times (24.587)^2
\]
\[
K.E. = 1000 \times 604.527 \approx 604,527 \, \text{J} 
\]

### Final answer:
The translational kinetic energy is approximately **6.25 x 10^5 J**, so the correct option is:

**6.25 x 10^5 [J]**

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Would you like more details or have any other questions? 

Here are 5 questions that extend this concept:

1. How does increasing the velocity affect the kinetic energy?
2. How would the kinetic energy change if the mass were doubled but the velocity remained the same?
3. What is the kinetic energy if the speed is increased to 60 mph?
4. How can potential energy be related to kinetic energy in this scenario?
5. What is the total energy of the system if the object is also 50 meters above the ground?

**Tip:** Kinetic energy increases with the square of velocity, so even small increases in speed can greatly increase energy.

What is the translational kinetic energy with a mass of 2000 [kg] travelling at the speed of 55 miles per hour?

Find the translational kinetic energy of an object with a mass of 2000 kg traveling at 55 miles per hour. Learn the conversion from mph to m/s and apply the formula KE = (1/2) * m * v^2 for a step-by-step solution. Suitable for students in Grades 10-12.

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