Question

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 stop at the first floor.

a. What is Zach’s weight before the elevator starts braking?

b. What is Zach’s weight while the elevator is braking?

Short answer

a. At rest $f_n=784N$

b. Upon brakingrole="math" localid="1647497386095" $f_n=1050.4N$

Step-by-step solution

Content Introduction

When the elevator is at rest or is moving at a constant speed, the net force that is acting on the Zach is zero. So that the apparent weight is equal to his real weight. $W_{appar}=mg$

Explanation (Part a)

When elevator is moving at a constant speed, Zach's apparent weight will be:

$$\begin{gathered} W_{appar}=m\times g \\ {W}_{appar}=80kg\times 9.8m/s^2 \\ {W}_{appar}=784N \end{gathered}$$

Explanation (Part b)

When the elevator is going downward and is slowing down, the net force is upward. Notice that the apparent weight is normal force.

$W_{appar}=ma+mg........\left(1\right)$

Now we will find the value of acceleration. We are given:

$$\begin{gathered} v_1=10m/s \\ {v}_f=0m/s \\ t=3.0s \end{gathered}$$

Therefore,

$$\begin{gathered} v_f=v_i+at \\ 0=10+(a\times 3) \\ a=-3.330m/s^2 \end{gathered}$$

We will make it positive because its direction is upward.

From (1)

$$\begin{gathered} W_{appar}=mg+ma \\ {W}_{appar}=m[g+a] \\ {W}_{appar}=80\times (9.8+3.33) \\ {W}_{appar}=1050.4N \end{gathered}$$