Question
A man with a mass of 80 kg is riding a motorcycle with a mass of 195 kg. The man and the motorcycle are traveling forward at 50 m/s.
Using the equation KE= ½mv2, calculate the kinetic energy of the man and the motorcycle.
A 243,750 J243,750 J
B 1,375,000 J1,375,000 J
C 1,900,880 J1,900,880 J
D 343,750 J
Using the equation KE= ½mv2, calculate the kinetic energy of the man and the motorcycle.
A 243,750 J243,750 J
B 1,375,000 J1,375,000 J
C 1,900,880 J1,900,880 J
D 343,750 J
Answers
GPT-4o mini
To calculate the kinetic energy of both the man and the motorcycle, we can use the formula for kinetic energy:
\[
KE = \frac{1}{2} mv^2
\]
where \( m \) is the mass and \( v \) is the velocity.
First, we need to find the total mass of the man and the motorcycle:
\[
m_{\text{total}} = m_{\text{man}} + m_{\text{motorcycle}} = 80 \, \text{kg} + 195 \, \text{kg} = 275 \, \text{kg}
\]
Next, we can substitute the total mass and the velocity into the kinetic energy formula:
\[
KE = \frac{1}{2} (275 \, \text{kg}) (50 \, \text{m/s})^2
\]
Calculating \( (50 \, \text{m/s})^2 \):
\[
(50 \, \text{m/s})^2 = 2500 \, \text{m}^2/\text{s}^2
\]
Now substituting this back into the kinetic energy equation:
\[
KE = \frac{1}{2} (275 \, \text{kg}) (2500 \, \text{m}^2/\text{s}^2)
\]
Calculating the product:
\[
275 \, \text{kg} \times 2500 \, \text{m}^2/\text{s}^2 = 687500 \, \text{kg}\cdot\text{m}^2/\text{s}^2
\]
Now, calculating the kinetic energy:
\[
KE = \frac{1}{2} (687500 \, \text{kg}\cdot\text{m}^2/\text{s}^2) = 343750 \, \text{J}
\]
Thus, the kinetic energy of the man and the motorcycle is:
\[
\boxed{343750 \, \text{J}}
\]
So the correct answer is **D: 343,750 J**.
\[
KE = \frac{1}{2} mv^2
\]
where \( m \) is the mass and \( v \) is the velocity.
First, we need to find the total mass of the man and the motorcycle:
\[
m_{\text{total}} = m_{\text{man}} + m_{\text{motorcycle}} = 80 \, \text{kg} + 195 \, \text{kg} = 275 \, \text{kg}
\]
Next, we can substitute the total mass and the velocity into the kinetic energy formula:
\[
KE = \frac{1}{2} (275 \, \text{kg}) (50 \, \text{m/s})^2
\]
Calculating \( (50 \, \text{m/s})^2 \):
\[
(50 \, \text{m/s})^2 = 2500 \, \text{m}^2/\text{s}^2
\]
Now substituting this back into the kinetic energy equation:
\[
KE = \frac{1}{2} (275 \, \text{kg}) (2500 \, \text{m}^2/\text{s}^2)
\]
Calculating the product:
\[
275 \, \text{kg} \times 2500 \, \text{m}^2/\text{s}^2 = 687500 \, \text{kg}\cdot\text{m}^2/\text{s}^2
\]
Now, calculating the kinetic energy:
\[
KE = \frac{1}{2} (687500 \, \text{kg}\cdot\text{m}^2/\text{s}^2) = 343750 \, \text{J}
\]
Thus, the kinetic energy of the man and the motorcycle is:
\[
\boxed{343750 \, \text{J}}
\]
So the correct answer is **D: 343,750 J**.