A block of mass 1 kg , moving along $x$ with speed $v_i=10 \mathrm{~m} / \mathrm{s}$ enters a rough region ranging from $x=0.1 \mathrm{~m}$ to $x=1.9 \mathrm{~m}$. The retarding force acting on the block in this range is $\mathrm{F}_{\mathrm{r}}=-\mathrm{kr} \mathrm{N}$, with k $=10 \mathrm{~N} / \mathrm{m}$. Then the final speed of the block as it crosses rough region is.

<p><strong>Work Done = Final Kinetic Energy - Initial Kinetic Energy</strong></p> <p>$W = \Delta K = K_f - K_i$</p> <p><p><strong>Calculate the Work Done by the Retarding Force</strong></p> <p>The problem gives a retarding force $F_r = -kr$. It seems 'r' is a typo for 'x', the position, since the constant $k$ has units of N/m. So, the force is a function of position: $F_r = -kx$.</p> <p>Because the force isn't constant, we need to find the work done by integrating the force over the distance the block travels in the rough region (from $x = 0.1 \text{ m}$ to $x = 1.9 \text{ m}$).</p> <p>$W = \int_{0.1}^{1.9} F_r \, dx = \int_{0.1}^{1.9} (-kx) \, dx$</p> <p>Plugging in $k = 10 \text{ N/m}$:</p> <p>$W = -10 \int_{0.1}^{1.9} x \, dx$</p> <p>$W = -10 \left[ \frac{x^2}{2} \right]_{0.1}^{1.9}$</p> <p>$W = -5 \left[ (1.9)^2 - (0.1)^2 \right]$</p> <p>$W = -5 [3.61 - 0.01] = -5(3.6)$</p> <p>$W = -18 \text{ J}$</p> <p>The negative sign just means the force did negative work, slowing the block down.</p></p> <p><p><strong>Calculate the Initial Kinetic Energy ($K_i$)</strong></p> <p>The initial kinetic energy is found using the formula $K_i = \frac{1}{2}mv_i^2$.</p> <p>Given:</p> <p><p>Mass $m = 1 \text{ kg}$</p></p> <p><p>Initial speed $v_i = 10 \text{ m/s}$</p></p> <p>$$K_i = \frac{1}{2}(1 \text{ kg})(10 \text{ m/s})^2 = \frac{1}{2}(100) = 50 \text{ J}$$</p></p> <p><p><strong>Use the Work-Energy Theorem to find the Final Speed ($v_f$)</strong></p> <p>Now we can put everything together into the Work-Energy Theorem equation: $W = K_f - K_i$.</p> <p>We know $W = -18 \text{ J}$ and $K_i = 50 \text{ J}$. The final kinetic energy is $K_f = \frac{1}{2}mv_f^2$.</p> <p>$-18 = \left(\frac{1}{2}(1)v_f^2\right) - 50$</p> <p>Now, let's solve for $v_f$:</p> <p>$-18 + 50 = \frac{1}{2}v_f^2$</p> <p>$32 = \frac{1}{2}v_f^2$</p> <p>$v_f^2 = 64$</p> <p>$v_f = \sqrt{64}$</p> <p>$v_f = 8 \text{ m/s}$</p></p> <p>The final speed of the block as it leaves the rough region is <strong>8 m/s</strong>.</p> <p>Therefore, the correct option is <strong>C</strong>.</p>