Question

⑥ a textbook is tossed down the hallway like a frisbee, hits the ground and slides through 7m to a stop in a time of 3.5 seconds.
a. draw a free body diagram of the book as its sliding along the floor.1
b. determine the coefficient of kinetic friction between the book and the floor.4

Explanation:

Step1: Find initial velocity

Use displacement formula: $d = \frac{v_i + v_f}{2}t$
Rearrange for $v_i$: $v_i = \frac{2d}{t} - v_f$
Substitute $d=7\,\text{m}$, $t=3.5\,\text{s}$, $v_f=0$:
$v_i = \frac{2\times7}{3.5} - 0 = 4\,\text{m/s}$

Step2: Calculate acceleration

Use velocity formula: $a = \frac{v_f - v_i}{t}$
Substitute values:
$a = \frac{0 - 4}{3.5} = -\frac{8}{7} \approx -1.14\,\text{m/s}^2$
(negative sign = deceleration)

Step3: Relate to friction force

Newton's 2nd law: $f_k = |ma|$, and $f_k = \mu_k F_N$
Since $F_N = mg$, substitute: $\mu_k mg = |ma|$
Cancel $m$, solve for $\mu_k$: $\mu_k = \frac{|a|}{g}$

Step4: Compute coefficient

Substitute $|a|=\frac{8}{7}\,\text{m/s}^2$, $g=9.8\,\text{m/s}^2$:
$\mu_k = \frac{8}{7\times9.8} = \frac{8}{68.6} \approx 0.117$

Answer:

a. Free body diagram:

• Upward normal force ($F_N$)
• Downward gravitational force ($F_g$)
• Leftward kinetic friction force ($f_k$) (opposing motion)

b. $\mu_k \approx 0.12$