Tom (65 kg) is standing on a scale in an elevator holding an apple. The elevator begins to accelerate downwards at 2.4 m/s2.
(a) What does the scale read during this acceleration?
(b) What is the acceleration of the apple with respect to Tom holding the apple?
2.4 m/s^2 (same as the elevator's acceleration)
am i right?
(c) Then as tom is accelerating up, he drops the apple. What is the acceleration of the apple with respect to you (i.e. an outsider looking on)?
(d) What is the acceleration of the apple with respect to Tom?
(e) What would the scale read if the elevator accelerated downward at 2.4 m/s2?
(a) N=m(g-a)= 65(9.8-2.4)= 481 Newtons
(b) 2.4 m²/s
(c) 9.8 m²/s
(d) a+g
To answer these questions, let's break down the problem step by step:
(a) What does the scale read during this acceleration?
To find the reading on the scale, we need to consider the forces acting on Tom while he's in the elevator. In this case, we have two forces: the force due to gravity (weight) and the force due to the acceleration of the elevator.
The weight of Tom can be calculated using the formula: weight = mass * acceleration due to gravity. Given that Tom's mass is 65 kg, and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate his weight as weight = 65 kg * 9.8 m/s^2 = 637 N.
Since the elevator is accelerating downwards, there would be an additional force acting on Tom in the opposite direction, opposing his weight. This force is equal to mass * acceleration, which is 65 kg * 2.4 m/s^2 = 156 N in this case.
The scale reading is the sum of these forces. Therefore, the scale would read 637 N - 156 N = 481 N during this acceleration.
(b) What is the acceleration of the apple with respect to Tom holding the apple?
The acceleration of the apple with respect to Tom is the same as the elevator's acceleration, which is given as 2.4 m/s^2.
(c) Then as Tom is accelerating up, he drops the apple. What is the acceleration of the apple with respect to you (i.e., an outsider looking on)?
From an outsider's perspective, the acceleration of the apple would be the same as the acceleration of the elevator before Tom dropped it, which is 2.4 m/s^2 downwards.
(d) What is the acceleration of the apple with respect to Tom?
Since Tom dropped the apple, it is no longer in his possession. Therefore, there is no longer any direct interaction between Tom and the apple, so the acceleration of the apple with respect to Tom would be zero.
(e) What would the scale read if the elevator accelerated downward at 2.4 m/s^2?
If the elevator accelerates downwards at 2.4 m/s^2, the scale would read the same as Tom's weight, which is 637 N. This is because the acceleration of the elevator matches the acceleration due to gravity, resulting in no additional force acting on Tom.