## Physics (10th Edition)

Published by Wiley

# Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 119: 112

#### Answer

The box travels a distance of $26.53cm$ along the incline.

#### Work Step by Step

1) The box's movement is opposed by 2 forces: - The component of the box's weight along the incline $mg\sin15$ - Kinetic friction $f_k=\mu_k F_N$ Since there is no vertical movement, $F_N=mg\cos15$ (the vertical component of the box's weight). Thus, $f_k=\mu_kmg\cos15$ We take the box's direction of motion to be on the positive side. According to Newton's 2nd Law: $$\sum F=-mg\sin15-\mu_kmg\cos15=ma$$ $$a=-g\sin15-\mu_kg\cos15$$ $$a=-9.8\sin15-0.18\times9.8\cos15=-4.24m/s^2$$ So the box decelerates at a rate $a=-4.24m/s^2$ 2) We have initial velocity $v_0=1.5$ and $v=0$. We can calculate the distance the box travels $$x=\frac{v^2-v_0^2}{2a}=0.2653m=26.53cm$$

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