Respuesta :
Answer:
a) W = 784 N
b) R = 1050.4 N
Explanation:
Given
Let us consider the upward direction positive
Mass of Zach m = 80 kg
initial velocity of the elevator u = -10 m/s
final velocity of the elevator v = 0 m/s
time taken to change velocity t = 3.0 s
Solution
a) Apparent weight before the braking
When the elevator is moving with a constant velocity, the only acceleration acts on Zach is the acceleration due to gravity
[tex]W = mg\\\\W = 80 \times 9.8 \\\\W = 784 N[/tex]
b) Apparent weight during the braking
When the elevator is breaking, its velocity changes
So its acceleration
[tex]a = \frac{\Delta v}{\Delta t} \\\\a = \frac{0-(-10)}{3} \\\\a = 3.33 m/s^2[/tex]
The positive sign indicates that the acceleration is acting upward
For upward acceleration, the apparent weight
[tex]R = m(g+a)\\\\R = 80 ( 9.8 + 3.33)\\\\R = 1050.4 N[/tex]
a. Zach’s apparent weight before the elevator starts braking is 784 Newton.
b. Zach’s apparent weight while the elevator is braking is 1050.4 Newton.
Given the following data:
- Mass = 80 kg
- Initial velocity = 10 m/s
- Time = 3 seconds
- Final velocity = 0 m/s (since the elevator brakes to a stop).
a. To find Zach’s apparent weight before the elevator starts braking:
- At constant velocity, Zach is being acted upon by an acceleration due to gravity.
Mathematically, the weight of an object is calculated by using the following formula;
[tex]W = mg[/tex]
Where:
- m is the mass of an object.
- g is the acceleration due to gravity.
[tex]Weight = 80(9.8)[/tex]
Weight = 784 Newton
b. To find Zach’s apparent weight while the elevator is braking:
First of all, we would determine the acceleration of the elevator.
[tex]Acceleration = \frac{V\; - \; U}{t} \\\\Acceleration = \frac{10\; - \; 0}{3}\\\\Acceleration = 3.33 \;m/s^2[/tex]
[tex]Apparent \; weight = m(a + g)\\\\Apparent \; weight = 80(3.33 + 9.8)\\\\Apparent \; weight = 80(13.13)[/tex]
Apparent weight = 1050.4 Newton.
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