How much more kinetic energy does a 6-kilogram bowling ball have when it is rolling at 16 mph (7.1 meters per second) than when it is rolling at 14 mph (6.2 meters per second)?

KE=12mv2

(1 point)
Responses

151.2 J
151.2 J

1.3 J
1.3 J

266.5 J
266.5 J

35.9 J

1 answer

To find out how much more kinetic energy the bowling ball has when rolling at 16 mph compared to 14 mph, we will use the kinetic energy formula:

\[ KE = \frac{1}{2}mv^2 \]

where:

  • \( m \) is the mass of the bowling ball (6 kg),
  • \( v \) is the velocity (in meters per second).

Step 1: Calculate the kinetic energy at 16 mph

Convert 16 mph to meters per second:

\[ 16 \text{ mph} \times \frac{1609.34 \text{ meters}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ seconds}} \approx 7.1 \text{ m/s} \]

Now, calculate the kinetic energy:

\[ KE_{16} = \frac{1}{2} \times 6 \text{ kg} \times (7.1 \text{ m/s})^2 \]

\[ KE_{16} = \frac{1}{2} \times 6 \times 50.41 \approx \frac{1}{2} \times 302.46 \approx 151.23 \text{ J} \]

Step 2: Calculate the kinetic energy at 14 mph

Convert 14 mph to meters per second:

\[ 14 \text{ mph} \times \frac{1609.34 \text{ meters}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ seconds}} \approx 6.2 \text{ m/s} \]

Now, calculate the kinetic energy:

\[ KE_{14} = \frac{1}{2} \times 6 \text{ kg} \times (6.2 \text{ m/s})^2 \]

\[ KE_{14} = \frac{1}{2} \times 6 \times 38.44 \approx \frac{1}{2} \times 230.64 \approx 115.32 \text{ J} \]

Step 3: Find the difference in kinetic energy

\[ \Delta KE = KE_{16} - KE_{14} \]

\[ \Delta KE = 151.23 \text{ J} - 115.32 \text{ J} \approx 35.91 \text{ J} \]

Conclusion

Therefore, the answer is approximately 35.9 J.