Question

A block of mass 1 kg, moving along x with speed \( v_i = 10 \) m/s enters a rough region ranging from \( x = 0.1 \) m to \( x = 1.9 \) m. The retarding force acting on the block in this range is \( F_r = -kx \) N, with \( k = 10 \) N/m. Then the final speed of the block as it crosses this rough region is

• A. 10 m/s
• B. 4 m/s
• C. 6 m/s
• D. 8 m/s

Hint:

Use the work-energy theorem to solve this problem. First, calculate the work done by the retarding force over the given distance by integrating the force with respect to displacement. Then, equate this work done to the change in kinetic energy of the block to find the final speed.

Answer 1

The Correct Option is D

The work performed by the retarding force on the block within the rough area is calculated as:\[W = \int_{x_i}^{x_f} F_r dx = \int_{0.1}^{1.9} (-kx) dx\]With \( k = 10 \) N/m, the equation becomes:\[W = \int_{0.1}^{1.9} (-10x) dx = -10 \left[ \frac{x^2}{2} \right]_{0.1}^{1.9}\]Evaluating the integral yields:\[W = -5 \left[ (1.9)^2 - (0.1)^2 \right] = -5 [3.61 - 0.01] = -5 [3.60] = -18 \, \text{J}\]The work-energy theorem states that the net work done on an object equals its change in kinetic energy:\[W_{net} = \Delta KE = KE_f - KE_i = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2\]Given a mass \( m = 1 \) kg and an initial speed \( v_i = 10 \) m/s, and considering the work done by the retarding force as the net work:\[-18 = \frac{1}{2} (1) v_f^2 - \frac{1}{2} (1) (10)^2\]Simplifying the equation:\[-18 = \frac{1}{2} v_f^2 - \frac{1}{2} (100)\]\[-18 = \frac{1}{2} v_f^2 - 50\]Rearranging to solve for \( v_f^2 \):\[\frac{1}{2} v_f^2 = 50 - 18 = 32\]\[v_f^2 = 64\]Therefore, the final speed is:\[v_f = \sqrt{64} = 8 \, \text{m/s}\]The block's final speed upon exiting the rough region is 8 m/s.