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Answer :
To solve the problem, we start with the formula for kinetic energy,
$$
KE = \frac{1}{2} m v^2,
$$
where $m$ is the mass and $v$ is the speed.
1. For the bowling ball with mass $m = 6$ kg rolling at $v_1 = 7.1$ m/s (16 mph), the kinetic energy is
$$
KE_1 = \frac{1}{2} \times 6 \times (7.1)^2.
$$
Evaluating the square and then the product,
$$
(7.1)^2 \approx 50.41,
$$
so
$$
KE_1 \approx 3 \times 50.41 = 151.23 \text{ J}.
$$
2. Similarly, when the bowling ball is rolling at $v_2 = 6.2$ m/s (14 mph),
$$
KE_2 = \frac{1}{2} \times 6 \times (6.2)^2.
$$
Calculating the square,
$$
(6.2)^2 \approx 38.44,
$$
we get
$$
KE_2 \approx 3 \times 38.44 = 115.32 \text{ J}.
$$
3. The difference in kinetic energy is then
$$
\Delta KE = KE_1 - KE_2 = 151.23 \text{ J} - 115.32 \text{ J} \approx 35.91 \text{ J}.
$$
Rounding this gives approximately $35.9$ J.
Thus, the bowling ball has about $\boxed{35.9\text{ J}}$ more kinetic energy when rolling at 16 mph compared to 14 mph.
$$
KE = \frac{1}{2} m v^2,
$$
where $m$ is the mass and $v$ is the speed.
1. For the bowling ball with mass $m = 6$ kg rolling at $v_1 = 7.1$ m/s (16 mph), the kinetic energy is
$$
KE_1 = \frac{1}{2} \times 6 \times (7.1)^2.
$$
Evaluating the square and then the product,
$$
(7.1)^2 \approx 50.41,
$$
so
$$
KE_1 \approx 3 \times 50.41 = 151.23 \text{ J}.
$$
2. Similarly, when the bowling ball is rolling at $v_2 = 6.2$ m/s (14 mph),
$$
KE_2 = \frac{1}{2} \times 6 \times (6.2)^2.
$$
Calculating the square,
$$
(6.2)^2 \approx 38.44,
$$
we get
$$
KE_2 \approx 3 \times 38.44 = 115.32 \text{ J}.
$$
3. The difference in kinetic energy is then
$$
\Delta KE = KE_1 - KE_2 = 151.23 \text{ J} - 115.32 \text{ J} \approx 35.91 \text{ J}.
$$
Rounding this gives approximately $35.9$ J.
Thus, the bowling ball has about $\boxed{35.9\text{ J}}$ more kinetic energy when rolling at 16 mph compared to 14 mph.
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Rewritten by : Jeany
