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Answer :
The magnitude of the braking force experienced by Zach is 3430 N (Option a).
The corect answer is (Option a).
To find the magnitude of the braking force experienced by Zach, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.
First, we need to find the acceleration of the elevator as it brakes to a stop. We can use the equation of motion [tex]\(v = u + at\),[/tex] where [tex]\(v\) i[/tex]s the final velocity (0 m/s, as the elevator stops), [tex]\(u\)[/tex] is the initial velocity [tex](12 m/s,[/tex] as the elevator descends), [tex]\(a\)[/tex]is the acceleration, and [tex]\(t\)[/tex]is the time taken to brake (2.9 s).
Rearranging the equation, we get[tex]\(a = \frac{v - u}{t}\).[/tex]
Substituting the given values, we find [tex]\(a = \frac{0 - 12}{2.9}\) m/s².[/tex]
This gives us the acceleration of the elevator during braking.
Next, we use Newton's second law [tex]\(F = ma\)[/tex] to find the braking force experienced by Zach.
Given Zach's mass [tex]\(m = 79\)[/tex] kg, and the calculated acceleration \[tex](a\)[/tex], we can calculate the braking force [tex]\(F\)[/tex].
Substituting the values, we get [tex]\(F = 79 \times \frac{-12}{2.9}\) N.[/tex]
Calculating this, we find [tex]\(F \approx -3430\) N.[/tex]
Since the magnitude of force cannot be negative, we take the absolute value to get [tex]\(F \approx 3430\) N.[/tex]
Thus, the magnitude of the braking force experienced by Zach is 3430 N. This force is exerted by the elevator's braking mechanism to bring it to a stop at the first floor, accounting for Zach's mass and the acceleration during braking.
The corect answer is (Option a).
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