Henry, whose mass is 95 kg, stands on a bathroom scale in an elevator. The scale reads 830 N for the first 3.0 s after the elevator starts moving, then 930 N for the next 3.0 s. What is the elevator's velocity 6.0 s after starting?

The elevator's velocity after 6 seconds is 3.189m/s

Explanation:

Using Newton's 2nd law:Fnet = ma

Mg - N = ma

Where m = mass

N = normal force

a = acceleration

g = acceleration due to gravity

Substituting the given values and finding acceleration

(95kg) (9.8m/s^2) - 830N = 95kg x a

a = (931 - 830) / 95

a = 101/95

a = 1.063m/s^2

After 3 seconds, the weight 0f the person is equal to the actual weight of the person, thus the elevator will be moving at a constant velocity.

Using kinematic equation:

V = u + at

Where V = final velocity

U = initial velocity = 0

a = acceleration

t = time

V = 0 + 1.063 * 3

V = 3.189m/s

This velocity does not change. The elevator travels the test of the time at the same velocity. Therefore, the velocity at 6 seconds = 3.189m/s