# Chapter 6 - Dynamics I: Motion Along a Line - Conceptual Questions - Page 160: 15

(a) The block of mass $2m$ slides a distance $d$. (b) The block slides a distance of $4d$.

#### Work Step by Step

(a) The blocks slide to a stop because of the force of kinetic friction. We can find the rate of deceleration $a_1$ for the block of mass $m$. $F_f = ma_1$ $mg~\mu_k = ma_1$ $a_1 = g~\mu_k$ We can find the rate of deceleration $a_2$ for the block of mass $2m$. $F_f = (2m)~a_2$ $(2m)~g~\mu_k = (2m)~a_2$ $a_2 = g~\mu_k$ Since the initial velocity is the same and the rate of deceleration is the same, then both blocks will slide the same distance $d$. The block of mass $2m$ slides a distance $d$. (b) $d_1 = \frac{0-(v_{0x})^2}{2a_1}$ We can find the distance that the block slides when the initial velocity is $2v_{0x}$. $d_2 = \frac{0-(2v_{0x})^2}{2a_2}$ $d_2 = 4\times \frac{0-(v_{0x})^2}{2a_1}$ $d_2 = 4~d_1$ The block slides a distance of $4d$.

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