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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.

The weight of a human doesn’t change significantly when you are on Earth. In situations that involve vertical acceleration the apparent weight of an object will change when the force acting on the object alters. As you accelerate upwards, a individual will appear heavier than they are. When they accelerate downwards, they will feel lighter.

Concepts and ReasonTo solve the problem, you will need to know the fundamental kinematic equation for motion for velocity as well as Newton’s second law.

Newton’s second law is used to calculate the apparent weight. Calculate the acceleration using the equation and the kinematic equation to calculate the elevator’s velocity. Calculate the apparent weight using the acceleration.

FundamentalsThe fundamental kinematic equation for motion states that an object’s final velocity is determined by its last velocity.

[katex]v = u + at[/katex]

Here, [katex]v[/katex]

Newton’s second law states that the force acting upon an object is:

[katex]F = ma[/katex]

Here, [katex]F[/katex]

Under acceleration due to gravity, the weight of an object is:

[katex]W = mg[/katex]

Here, [katex]W[/katex]

Before the elevator begins moving, the apparent weight of the passenger is:

[katex]W = mg[/katex]

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

The fundamental kinematic equation for motion states that the final velocity of a passenger is.

[katex]v = u + at[/katex]

The passenger’s initial velocity will be zero because it starts at rest.

[katex]u = 0[/katex]

[katex]\begin{array}{c}\\v = 0 + at\\\\a = \frac{v}{t}\\\end{array}[/katex]

When a passenger is moving upward, the net force that acts on him is:

[katex]F = {W_a} – W[/katex]

Here, [katex]F[/katex]

Newton’s second law is that the passenger is the net force.

[katex]F = ma[/katex]

[katex]F = \frac{{mv}}{t}[/katex]

Under acceleration due to gravity, the passenger’s weight is:

[katex]W = mg[/katex]

Rewrite the equation to calculate net force $equaTag13

The passenger’s apparent weight is

[katex]{W_a} = F + W[/katex]

Substitute [katex]\frac{{mv}}{t}[/katex]

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

[katex]\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}[/katex]

738N is the apparent weight of the passenger as the elevator speeds up.

The elevator slows down and stops accelerating upward when it reaches the cruising speed. Only gravity’s acceleration can cause acceleration to the passenger.

The passenger’s apparent weight is

[katex]W = mg[/katex]

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

After the elevator reaches cruising speed, the apparent weight of the passenger is [katex]588\,{\rm{N}}[/katex]

Ans.The passenger’s apparent weight before the elevator begins moving is 588 N

738N is the apparent weight of the passenger as the elevator speeds up.

After the elevator reaches cruising speed, the apparent weight of the passenger is 588 N