# Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems: 15

At t = 1.0 s, the person's apparent weight is 1040 N At t = 5.0 s, the person's apparent weight is 735 N At t = 9.0 s, the person's apparent weight is 585 N

#### Work Step by Step

Let $F_N$ be the normal force of the elevator floor pushing up on the person. Note that $F_N$ is equal in magnitude to the person's apparent weight. The acceleration $a_y$ is equal to the slope of the $v_y$ versus time graph. At t = 1.0 s: $a_y = \frac{\Delta v_y}{\Delta t}$ $a_y = \frac{8.0~m/s-0}{2.0~s-0}$ $a_y = 4.0~m/s^2$ We can use $a_y$ to find the apparent weight. $\sum F = ma_y$ $F_N- mg = ma_y$ $F_N = m(g+a_y)$ $F_N = (75~kg)(9.80~m/s^2+ 4.0~m/s^2)$ $F_N = 1040~N$ The person's apparent weight is 1040 N At t = 5.0 s: $a_y = \frac{\Delta v_y}{\Delta t}$ $a_y = \frac{8.0~m/s-8.0~m/s}{6.0~s-2.0~s}$ $a_y = 0$ We can use $a_y$ to find the apparent weight. $\sum F = ma_y$ $F_N- mg = ma_y$ $F_N = m(g+a_y)$ $F_N = (75~kg)(9.80~m/s^2+ 0)$ $F_N = 735~N$ The person's apparent weight is 735 N At t = 9.0 s: $a_y = \frac{\Delta v_y}{\Delta t}$ $a_y = \frac{0-8.0~m/s}{10.0~s-6.0~s}$ $a_y = -2.0~m/s^2$ We can use $a_y$ to find the apparent weight. $\sum F = ma_y$ $F_N- mg = ma_y$ $F_N = m(g+a_y)$ $F_N = (75~kg)(9.80~m/s^2-2.0~m/s^2)$ $F_N = 585~N$ The person's apparent weight is 585 N

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