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Answer :
To solve the problem, we start by finding the acceleration using the formula:
[tex]$$
a = \frac{v_f - v_i}{t}
$$[/tex]
where
- [tex]$v_i = 10 \, \text{m/s}$[/tex] is the initial velocity,
- [tex]$v_f = 16 \, \text{m/s}$[/tex] is the final velocity, and
- [tex]$t = 10 \, \text{s}$[/tex] is the time over which the acceleration occurs.
Substituting the values, we have:
[tex]$$
a = \frac{16 \, \text{m/s} - 10 \, \text{m/s}}{10 \, \text{s}} = \frac{6 \, \text{m/s}}{10 \, \text{s}} = 0.6 \, \text{m/s}^2.
$$[/tex]
Next, we apply Newton's second law of motion, which is given by:
[tex]$$
F = m \cdot a
$$[/tex]
where
- [tex]$m = 280 \, \text{kg}$[/tex] is the mass of the snowmobile (including Sully), and
- [tex]$a = 0.6 \, \text{m/s}^2$[/tex] is the acceleration we just calculated.
Now, calculate the force:
[tex]$$
F = 280 \, \text{kg} \times 0.6 \, \text{m/s}^2 = 168 \, \text{N}.
$$[/tex]
Thus, the force exerted by the snowmobile to accelerate is [tex]$\boxed{168\,\text{N}}$[/tex], which corresponds to answer choice B.
[tex]$$
a = \frac{v_f - v_i}{t}
$$[/tex]
where
- [tex]$v_i = 10 \, \text{m/s}$[/tex] is the initial velocity,
- [tex]$v_f = 16 \, \text{m/s}$[/tex] is the final velocity, and
- [tex]$t = 10 \, \text{s}$[/tex] is the time over which the acceleration occurs.
Substituting the values, we have:
[tex]$$
a = \frac{16 \, \text{m/s} - 10 \, \text{m/s}}{10 \, \text{s}} = \frac{6 \, \text{m/s}}{10 \, \text{s}} = 0.6 \, \text{m/s}^2.
$$[/tex]
Next, we apply Newton's second law of motion, which is given by:
[tex]$$
F = m \cdot a
$$[/tex]
where
- [tex]$m = 280 \, \text{kg}$[/tex] is the mass of the snowmobile (including Sully), and
- [tex]$a = 0.6 \, \text{m/s}^2$[/tex] is the acceleration we just calculated.
Now, calculate the force:
[tex]$$
F = 280 \, \text{kg} \times 0.6 \, \text{m/s}^2 = 168 \, \text{N}.
$$[/tex]
Thus, the force exerted by the snowmobile to accelerate is [tex]$\boxed{168\,\text{N}}$[/tex], which corresponds to answer choice B.
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