Question

. A 47 kg child is standing on a scale in an elevator, use this for each of the following cases:

A) What does the scale read while the elevator is traveling down while slowing at 2.5 m/s2?

B) What is the acceleration is the scale reads 555 N?

Answer 1

(A) Let’s apply the Newton’s Second Law of Motion:

$$\sum F_y=ma_y,$$ $$N-mg=-ma,$$

here, $N$ is the normal force directed upward, $mg$ is the force of gravity directed downward, $a$ is the acceleration of an elevator (since the acceleration of an elevator directed downward it will be with sign minus).

Then, we get:

$$N=m(g-a)=47\ kg\cdot(9.8\ \dfrac{m}{s^2}-2.5\ \dfrac{m}{s^2})=343.1\ N.$$

(B) Let’s apply the Newton’s Second Law of Motion:

$$\sum F_y=ma_y,$$ $$N-mg=ma,$$ $$a=\dfrac{N-mg}{m},$$ $$a=\dfrac{555\ N-47\ kg\cdot 9.8\ \dfrac{m}{s^2}}{47\ kg}=2.0\ \dfrac{m}{s^2}.$$

The sign plus means that the elevator moves upward.

Answer:

(A) $N=343.1\ N.$

(B) $a=2.0\ \dfrac{m}{s^2},$ upward.