A 95.0-kg person stands on a scale in an elevator. What is the apparent weight when the elevator is (a) accelerating upward with an acceleration of 1.80 m/s2 , (b) moving upward at a constant speed, and (c) accelerating downward with an acceleration of 1.30 m/s2 ?

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madeleinejames20 · Accepted Answer

To solve this problem, we need to consider the forces acting on the person in each situation.(a) When the elevator is accelerating upward with an acceleration of 1.80 m/s², the apparent weight can be determined using the equation:Fnet=m·awhere Fnet is the net force acting on the person, m is the mass of the person, and a is the acceleration.In this case, the net force is the difference between the gravitational force and the normal force (which is the reading on the scale). The gravitational force is given by:Fgravity=m·gwhere g is the acceleration due to gravity.Since the person is in an elevator, the normal force is the apparent weight of the person. Thus, we can rewrite the equation as:Fgravity−Apparent&amp;nbsp;weight=m·aRearranging the equation to solve for the apparent weight:Apparent&amp;nbsp;weight=Fgravity−m·aSubstituting the given values:Apparent&amp;nbsp;weight=(m·g)−(m·a)Apparent&amp;nbsp;weight=m·(g−a)Plugging in the values:Apparent&amp;nbsp;weight=95.0kg·(9.8m/s²−1.80m/s²)Apparent&amp;nbsp;weight=95.0kg·7.00m/s²Apparent&amp;nbsp;weight=665.0NTherefore, the apparent weight when the elevator is accelerating upward with an acceleration of 1.80 m/s² is 665.0 N.(b) When the elevator is moving upward at a constant speed, there is no acceleration. In this case, the net force is zero, which means the apparent weight is equal to the gravitational force.Apparent&amp;nbsp;weight=Fgravity=m·gSubstituting the given values:Apparent&amp;nbsp;weight=95.0kg·9.8m/s²Apparent&amp;nbsp;weight=931.0NTherefore, the apparent weight when the elevator is moving upward at a constant speed is 931.0 N.(c) When the elevator is accelerating downward with an acceleration of 1.30 m/s², we follow the same steps as in part (a):Apparent&amp;nbsp;weight=Fgravity−m·aApparent&amp;nbsp;weight=m·(g−a)Substituting the given values:Apparent&amp;nbsp;weight=95.0kg·(9.8m/s²−1.30m/s²)Apparent&amp;nbsp;weight=95.0kg·8.50m/s²Apparent&amp;nbsp;weight=807.5NTherefore, the apparent weight when the elevator is accelerating downward with an acceleration of 1.30 m/s² is 807.5 N.

A 95.0-kg person stands on a scale in

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