Question

A student takes the elevator up to the fourth floor to see her favorite physics instructor. She stands on the floor of the elevator, which is horizontal. Both the student and the elevator are solid objects, and they both accelerate upward at 5.10 m/s2. This acceleration only occurs briefly at the beginning of the ride up. Her mass is 81.5 kg. What is the normal force exerted by the floor of the elevator on the student during her brief acceleration? (Assume the j direction is upward.)

Answer 1

To calculate the normal force exerted by the elevator during acceleration, sum the student's weight and the additional force from the elevator's upward acceleration. The student's weight is 798.7 N, and the additional force due to acceleration is 415.65 N, resulting in a normal force of 1214.35 N.

Step-by-step explanation:

To calculate the normal force exerted by the floor of the elevator on the student, we must consider both the gravitational force acting on the student and the additional force resulting from the upward acceleration of the elevator. The gravitational force, also known as weight, is the product of the student's mass and the acceleration due to gravity (g = 9.8 m/s2). The additional force is the product of the student's mass and the elevator's acceleration (a).

Normal force (N) calculation:

• Weight (W) = mass (m) × acceleration due to gravity (g)
• W = 81.5 kg × 9.8 m/s2 = 798.7 N
• Additional force due to elevator acceleration = mass (m) × elevator acceleration (a)
• Additional force = 81.5 kg × 5.10 m/s2 = 415.65 N
• Total normal force = Weight + Additional force
• N = 798.7 N + 415.65 N = 1214.35 N

Therefore, the normal force exerted on the student during her brief acceleration is 1214.35 Newtons (N).

Answer 2

The normal force exerted by the floor of the elevator on the student during her brief acceleration is 1214.35 N.

Step-by-step explanation:

Given that,

Acceleration = 5.10 m/s²

Mass of instructor = 81.5 kg

Normal force is in upward direction and weight of the student in downward direction

So, acceleration is in upward direction

We need to calculate the normal force exerted by the floor of the elevator on the student during her brief acceleration

Using formula of force

`F_(N)-mg=ma`

`N_(N)=m(g+a)`

Put the value into the formula

`F_(N)=81.5(9.8+5.10)`

`F_(N)=1214.35\ N`

Hence, The normal force exerted by the floor of the elevator on the student during her brief acceleration is 1214.35 N.