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Answer :
To solve the problem, we start with the formula that relates force, mass, and acceleration:
$$ F = m \cdot a. $$
First, we calculate the acceleration. The acceleration is given by the formula
$$ a = \frac{v_f - v_i}{t}, $$
where:
- $v_i$ is the initial velocity,
- $v_f$ is the final velocity, and
- $t$ is the time taken.
Given that the snowmobile's initial velocity is $10 \, \text{m/s}$, the final velocity is $16 \, \text{m/s}$, and the time interval is $10 \, \text{s}$, the acceleration is
$$ a = \frac{16 \, \text{m/s} - 10 \, \text{m/s}}{10 \, \text{s}} = \frac{6 \, \text{m/s}}{10 \, \text{s}} = 0.6 \, \text{m/s}^2. $$
Next, we use the mass of the snowmobile (including Sully), which is $280 \, \text{kg}$, in the force equation:
$$ F = 280 \, \text{kg} \times 0.6 \, \text{m/s}^2 = 168 \, \text{N}. $$
Thus, the force exerted by the snowmobile to accelerate is
$$ 168 \, \text{N}. $$
Therefore, the correct answer is:
B. 168 N.
$$ F = m \cdot a. $$
First, we calculate the acceleration. The acceleration is given by the formula
$$ a = \frac{v_f - v_i}{t}, $$
where:
- $v_i$ is the initial velocity,
- $v_f$ is the final velocity, and
- $t$ is the time taken.
Given that the snowmobile's initial velocity is $10 \, \text{m/s}$, the final velocity is $16 \, \text{m/s}$, and the time interval is $10 \, \text{s}$, the acceleration is
$$ a = \frac{16 \, \text{m/s} - 10 \, \text{m/s}}{10 \, \text{s}} = \frac{6 \, \text{m/s}}{10 \, \text{s}} = 0.6 \, \text{m/s}^2. $$
Next, we use the mass of the snowmobile (including Sully), which is $280 \, \text{kg}$, in the force equation:
$$ F = 280 \, \text{kg} \times 0.6 \, \text{m/s}^2 = 168 \, \text{N}. $$
Thus, the force exerted by the snowmobile to accelerate is
$$ 168 \, \text{N}. $$
Therefore, the correct answer is:
B. 168 N.
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