Question

23. A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 591 N. As the elevator later stops, the scale reading is 391 N Assuming the magnitude of the acceleration is the same during starting and stopping, determine (a) the weight of the person, (b) the person’s mass, and (c) the acceleration of the elevator.

Short answer

(a) The weight of the person 491 N.

(b) The person’s mass 50.1 kg.

(c) The acceleration of the elevator 2.00 m/s².

Step-by-step solution

Determining the weight of the person

(a)

The scale measures the upward normal force imposed on the passenger by the floor. During upward acceleration, the highest force is generated (when starting an upward trip or ending a downward trip). With downward acceleration, the minimal normal force occurs. For each unique circumstance

$\sum F_y=m{a}_y$

becomes for starting

$+591N-mg=+ma$

and for stopping

$+391N-mg=-ma$

The magnitude of the acceleration is represented by a.

To cancel out a and solve for mg, combine these two simultaneous equations

$\begin{array}{c}+591N-mg+391N-mg=0\\ 982N-2mg=0\\ F_g=mg\\ =\frac{982N}{2}\\ =491N\end{array}$

Determining the mass of the person

(b)

From the definition of weight

$\begin{array}{c}m=\frac{F_g}{g}\\ =\frac{491N}{9.80m/s^2}\\ =50.1kg\end{array}$

Determining the acceleration of the elevator

(c)

Substituting back gives

$\begin{array}{c}+591N-491N=(50.1kg)a\\ a=\frac{100N}{50.1kg}\\ =2.00m/s^2\end{array}$