High School

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)?

\[ KE = \frac{1}{2} m v^2 \]

A. 151.2 J
B. 1.3 J
C. 266.5 J
D. 35.9 J

Answer :

To find the additional kinetic energy when the bowling ball is rolling at 16 mph (7.1 m/s) compared to 14 mph (6.2 m/s), we start with the formula for kinetic energy:

$$
KE = \frac{1}{2} m v^2
$$

where:
- $m$ is the mass of the ball,
- $v$ is its speed.

Given:
- Mass, $m = 6\ \text{kg}$
- Speed at 16 mph, $v_1 = 7.1\ \text{m/s}$
- Speed at 14 mph, $v_2 = 6.2\ \text{m/s}$

**Step 1: Compute the squares of the speeds.**

For $v_1$:

$$
v_1^2 = (7.1)^2 \approx 50.41\ \text{m}^2/\text{s}^2
$$

For $v_2$:

$$
v_2^2 = (6.2)^2 \approx 38.44\ \text{m}^2/\text{s}^2
$$

**Step 2: Compute the kinetic energy at each speed.**

For $v_1$:

$$
KE_1 = \frac{1}{2} \times 6 \times 50.41 \approx 3 \times 50.41 \approx 151.23\ \text{J}
$$

For $v_2$:

$$
KE_2 = \frac{1}{2} \times 6 \times 38.44 \approx 3 \times 38.44 \approx 115.32\ \text{J}
$$

**Step 3: Calculate the difference in kinetic energy.**

$$
\Delta KE = KE_1 - KE_2 \approx 151.23\ \text{J} - 115.32\ \text{J} \approx 35.91\ \text{J}
$$

Rounded to one decimal place, the additional kinetic energy is approximately:

$$
35.9\ \text{J}
$$

Thus, the 6-kilogram bowling ball has approximately $35.9$ joules more kinetic energy at 16 mph than at 14 mph.

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