Q74P - University Physics with Modern Physics

Question

An elevator is moving upward at a constant speed of $\mathbf{2}\mathbf{.}\mathbf{50}\mathit{m}\mathbf{/}\mathit{s}$ . A bolt in the elevator ceiling above the elevator floor works loose and falls. (a) How long does it take for the bolt to fall to the elevator floor? What is the speed of the bolt just as it hits the elevator floor (b) according to an observer in the elevator? (c) According to an observer standing on one of the floor landings of the building? (d) According to the observer in part (c), what distance did the bolt travel between the ceiling and the floor of the elevator?

Step-by-Step Solution

Answer

1. $t=0.782\text{s}$
   
2. $-7.67m/s$
   
3. $-5.17m/s$
   
4. $s=-1.04\text{m}$

Hence, $t=0.782\text{s}$ is the time taking in falling of bolt the elevator floor, $-7.67m/s$ is the velocity of bolt relative to an observer in the elevator, $-5.17m/s$ is the velocity of bolt related to earth, $s=-1.04\text{m}$ is the distance the bolt travel between the ceiling and the floor of the elevator.

Step 1: Newton’s law

The velocity contains two components, one is horizontal component and another one is vertical.

 

According to the newton’s laws of motion,

$s=ut+\frac{1}{2}a{t}^2$

Where, s, u, t, a, are displacement, initial velocity, time and acceleration.

Step 2: Given

Elevator =$2.50m/s$

Bolt height= $3.00m$

Step 3: (a) Time taking in falling of bolt the elevator floor

From the vertical motion, initial velocity 

 

The position of the volt

$\text{3.00m+}\left(2.50\mathrm{m}/\mathrm{s}\right)\text{t-}\left(1/2\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right){\text{t}}^2$

And the position of the floor is,

$\left(2.50\mathrm{m}/\mathrm{s}\right)t$

On equating the both,

$$\begin{gathered} \text{3.00m}=\left(1/2\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right){\mathrm{t}}^2 \\ \mathrm{t}=0.782\mathrm{s} \end{gathered}$$

Step 4: (b) The velocity of bolt relative to an observer in the elevator

The velocity of bolt relative to the earth,

$\left(\text{2.50m/s}\right)-\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(0.782\text{s}\right)=-5.17\mathrm{m}/\mathrm{s}$

So, the velocity of bolt relative to an observer in the elevator

$-5.17\mathrm{m}/\mathrm{s}-\text{2.50m/s=}-7.67\mathrm{m}/\mathrm{s}$

Step 5: (c) Velocity of bolt related to earth

The velocity of bolt relative to the earth,

$-5.17\mathrm{m}/\mathrm{s}-\text{2.50m/s=}-7.67\mathrm{m}/\mathrm{s}$

Step 6: (d) distance did the bolt travel between the ceiling and the floor of the elevator

The distance traveled by bolt relative to the earth,

$$\begin{gathered} s=\left(\text{2.50m/s}\right)\mathrm{t}-\frac{1}{2}\left(9.80\mathrm{m}/{\mathrm{s}}^2\right){\mathrm{t}}^2 \\ \mathrm{s}=\left(\text{2.50m/s}\right)\left(0.782\text{s}\right)-\frac{1}{2}\left(9.80\mathrm{m}/{\mathrm{s}}^2\right){\left(0.782\text{s}\right)}^2 \\ \mathrm{s}=-1.04\text{m} \end{gathered}$$

Hence, $t=0.782\text{s}$is the time taking in falling of bolt the elevator floor, $-7.67\mathrm{m}/\mathrm{s}$ is the velocity of bolt relative to an observer in the elevator, $-5.17m/s$ is the velocity of bolt related to earth, data-custom-editor="chemistry" $s=-1.04\text{m}$ is the distance the bolt travel between the ceiling and the floor of the elevator.