Answer :
We start with the kinetic energy formula:
$$
KE = \frac{1}{2} m v^2,
$$
where $m$ is the mass and $v$ is the speed.
1. For the bowling ball with mass $m = 6\text{ kg}$ rolling at $v = 7.1\text{ m/s}$, the kinetic energy is calculated as:
$$
KE_1 = \frac{1}{2} \times 6 \times (7.1)^2.
$$
Evaluating the expression:
$$
(7.1)^2 \approx 50.41 \quad \text{and} \quad \frac{1}{2} \times 6 = 3,
$$
so
$$
KE_1 \approx 3 \times 50.41 = 151.23\text{ J}.
$$
2. For the ball rolling at $v = 6.2\text{ m/s}$, the kinetic energy is:
$$
KE_2 = \frac{1}{2} \times 6 \times (6.2)^2.
$$
Calculating further:
$$
(6.2)^2 \approx 38.44 \quad \text{and} \quad \frac{1}{2} \times 6 = 3,
$$
so
$$
KE_2 \approx 3 \times 38.44 = 115.32\text{ J}.
$$
3. The increase in kinetic energy when the bowling ball goes from $6.2\text{ m/s}$ to $7.1\text{ m/s}$ is found by the difference:
$$
\Delta KE = KE_1 - KE_2 \approx 151.23\text{ J} - 115.32\text{ J} = 35.91\text{ J}.
$$
Thus, the bowling ball has approximately $35.9\text{ J}$ more kinetic energy when rolling at $7.1\text{ m/s}$ compared to $6.2\text{ m/s}$.
The correct answer is $\boxed{35.9\text{ J}}$.
$$
KE = \frac{1}{2} m v^2,
$$
where $m$ is the mass and $v$ is the speed.
1. For the bowling ball with mass $m = 6\text{ kg}$ rolling at $v = 7.1\text{ m/s}$, the kinetic energy is calculated as:
$$
KE_1 = \frac{1}{2} \times 6 \times (7.1)^2.
$$
Evaluating the expression:
$$
(7.1)^2 \approx 50.41 \quad \text{and} \quad \frac{1}{2} \times 6 = 3,
$$
so
$$
KE_1 \approx 3 \times 50.41 = 151.23\text{ J}.
$$
2. For the ball rolling at $v = 6.2\text{ m/s}$, the kinetic energy is:
$$
KE_2 = \frac{1}{2} \times 6 \times (6.2)^2.
$$
Calculating further:
$$
(6.2)^2 \approx 38.44 \quad \text{and} \quad \frac{1}{2} \times 6 = 3,
$$
so
$$
KE_2 \approx 3 \times 38.44 = 115.32\text{ J}.
$$
3. The increase in kinetic energy when the bowling ball goes from $6.2\text{ m/s}$ to $7.1\text{ m/s}$ is found by the difference:
$$
\Delta KE = KE_1 - KE_2 \approx 151.23\text{ J} - 115.32\text{ J} = 35.91\text{ J}.
$$
Thus, the bowling ball has approximately $35.9\text{ J}$ more kinetic energy when rolling at $7.1\text{ m/s}$ compared to $6.2\text{ m/s}$.
The correct answer is $\boxed{35.9\text{ J}}$.