Answered

Zach, whose mass is 80 kg, is in an elevator descending at

10 m/s. The elevator takes 3.0 s to brake to a stopat the first

floor.

AWhat is Zach’s apparent weight before theelevator starts

braking?

BWhat is Zach’s apparent weight while theelevator is braking?

Verified Answer :

Chlidonias

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

\(W = mg\\\\W = 80 \times  9.8 \\\\W = 784 N\)

b) Apparent weight during the braking

When the elevator is breaking, its velocity changes

So its acceleration

\(a = \frac{\Delta v}{\Delta t} \\\\a = \frac{0-(-10)}{3} \\\\a = 3.33 m/s^2\)

The positive sign indicates that the acceleration is acting upward

For upward acceleration, the apparent weight

\(R = m(g+a)\\\\R = 80 ( 9.8 + 3.33)\\\\R = 1050.4 N\)

Lanuel

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;

\(W = mg\)

Where:

  • m is the mass of an object.
  • g is the acceleration due to gravity.

\(Weight = 80(9.8)\)

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.

\(Acceleration = \frac{V\; - \; U}{t} \\\\Acceleration = \frac{10\; - \; 0}{3}\\\\Acceleration = 3.33 \;m/s^2\)

\(Apparent \; weight = m(a + g)\\\\Apparent \; weight = 80(3.33 + 9.8)\\\\Apparent \; weight = 80(13.13)\)

Apparent weight = 1050.4 Newton.

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