Question

It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor.

What is the passenger's apparent weight before the elevator starts moving?

Express your answer using two significant figures.

What is the passenger's apparent weight while the elevator is speeding up?

Express your answer using two significant figures.

What is the passenger's apparent weight after the elevator reaches its cruising speed?

Express your answer using two significant figures.

Answer 1

Concepts and reason

The concepts required to solve the problem are the fundamental kinematic equation of motion for velocity and Newton’s second law of motion.

To calculate the apparent weight, use Newton’s second law of motion. Use the kinematic equation for velocity of the elevator and calculate the acceleration from the equation. Use the acceleration to calculate the apparent weight.

Fundamentals

According to the fundamental kinematic equation of motion, the final velocity of an object is,

$v = u + at$

Here, $v$ is the final velocity of the object, $u$ is the initial velocity of the object, $a$ is the acceleration of the object and $t$ is the time.

According to Newton’s second law, the force acting on an object is:

$F = ma$

Here, $F$ is the force, $m$ is the mass of the object and $a$ is the acceleration of the object.

The weight of an object under acceleration due to gravity is:

$W = mg$

Here, $W$ is the weight of the object, $m$ is the mass of the object, and $g$ is the acceleration due to gravity.

The apparent weight of the passenger before the elevator starts moving is:

$W = mg$

Substitute $60\,{\rm{kg}}$ for $m$ and $9.8\,{\rm{m/}}{{\rm{s}}^2}$ for $g$ to calculate the apparent weight of the passenger.

$\begin{array}{c}\\W = \left( {60\,{\rm{kg}}} \right)\left( {9.8\,{\rm{m/}}{{\rm{s}}^2}} \right)\\\\ = 588\,{\rm{N}}\\\end{array}$

According to the fundamental kinematic equation of motion, the final velocity of the passenger is,

$v = u + at$

Since the passenger starts from rest, the initial velocity of the passenger will be zero.

$u = 0$

Substitute $0$ for$u$ in the equation$v = u + at$ and solve for$a$.

$\begin{array}{c}\\v = 0 + at\\\\a = \frac{v}{t}\\\end{array}$

The net force acting on the passenger when he is moving upward is:

$F = {W_a} - W$

Here, $F$ is the net force on the passenger, $W$ is the weight of the passenger under acceleration due to gravity, and ${W_a}$ is the apparent weight of the passenger when the elevator moves upward.

According to Newton’s second law of motion, the net force acting on the passenger is,

$F = ma$

Substitute $\frac{v}{t}$ for $a$ in the above expression to rewrite F.

$F = \frac{{mv}}{t}$

The weight of the passenger under acceleration due to gravity is,

$W = mg$

Rewrite the equation for net force $F = {W_a} - W$in terms of apparent weight of the passenger.

The apparent weight of the passenger is,

${W_a} = F + W$

Substitute $\frac{{mv}}{t}$ for $F$and $mg$ for $W$ in ${W_a} = F + W$ in the above expression to rewrite ${W_a}$.

$\begin{array}{c}\\{W_a} = \frac{{mv}}{t} + mg\\\\ = m\left( {\frac{v}{t} + g} \right)\\\end{array}$

Substitute $60\,{\rm{kg}}$ for $m$, $10\,{\rm{m/s}}$ for $v$, $4.0\,{\rm{s}}$ for $t$, and $9.8\,{\rm{m/}}{{\rm{s}}^2}$ for acceleration due to gravity to calculate ${W_a}$.

$\begin{array}{c}\\{W_a} = \left( {60\,{\rm{kg}}} \right)\left( {\frac{{10\,{\rm{m/s}}}}{{4.0\,{\rm{s}}}} + 9.8\,{\rm{m/}}{{\rm{s}}^2}} \right)\\\\ = 738\,{\rm{N}}\\\end{array}$

The apparent weight of the passenger while the elevator is speeding up is $738\,{\rm{N}}$.

When the elevator reaches the cruising speed, the elevator stops accelerating upward and moves with the constant speed. The only acceleration acting on the passenger is the acceleration due to gravity.

The apparent weight of the passenger is,

$W = mg$

Substitute $60\,{\rm{kg}}$ for $m$ and $9.8\,{\rm{m/}}{{\rm{s}}^2}$ for $g$ to calculate the apparent weight of the passenger.

$\begin{array}{c}\\W = \left( {60\,{\rm{kg}}} \right)\left( {9.8\,{\rm{m/}}{{\rm{s}}^2}} \right)\\\\ = 588\,{\rm{N}}\\\end{array}$

The apparent weight of the passenger after the elevator reaches the cruising speed is $588\,{\rm{N}}$.

Ans:

The apparent weight of the passenger before the elevator starts moving is$588\,{\rm{N}}$.

The apparent weight of the passenger while the elevator is speeding up is $738\,{\rm{N}}$.

The apparent weight of the passenger after the elevator reaches the cruising speed is $588\,{\rm{N}}$.