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

266.5 J
266.5 J

35.9 J
35.9 J

151.2 J
151.2 J

1.3 J
1.3 J

1 answer

To find the difference in kinetic energy of the bowling ball at the two different speeds, we will use the formula for kinetic energy:

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

where \( m \) is the mass (6 kg) and \( v \) is the velocity.

First, we find the kinetic energy at 16 mph (7.1 m/s):

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

Calculating the velocity squared:

\[ (7.1 , \text{m/s})^2 = 50.41 , \text{m}^2/\text{s}^2 \]

Now, calculate the kinetic energy:

\[ KE_{16} = \frac{1}{2} \cdot 6 \cdot 50.41 = 3 \cdot 50.41 = 151.23 , \text{J} \]

Next, we find the kinetic energy at 14 mph (6.2 m/s):

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

Calculating the velocity squared:

\[ (6.2 , \text{m/s})^2 = 38.44 , \text{m}^2/\text{s}^2 \]

Calculate the kinetic energy:

\[ KE_{14} = \frac{1}{2} \cdot 6 \cdot 38.44 = 3 \cdot 38.44 = 115.32 , \text{J} \]

Now, find the difference in kinetic energy:

\[ \Delta KE = KE_{16} - KE_{14} = 151.23 , \text{J} - 115.32 , \text{J} = 35.91 , \text{J} \]

Thus, the difference in kinetic energy is approximately \( 35.9 , \text{J} \).

The correct answer is:

35.9 J