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 done by the retarding force on the block as it moves through the rough region is given by: \[ W = \int_{x_i}^{x_f} F_r dx = \int_{0.1}^{1.9} (-kx) dx \] Substituting \( k = 10 \) N/m: \[ W = \int_{0.1}^{1.9} (-10x) dx = -10 \left[ \frac{x^2}{2} \right]_{0.1}^{1.9} \] \[ W = -5 \left[ (1.9)^2 - (0.1)^2 \right] = -5 [3.61 - 0.01] = -5 [3.60] = -18 \, \text{J} \] The work-energy theorem states that the net work done on an object is equal to the change in its 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 mass \( m = 1 \) kg and initial speed \( v_i = 10 \) m/s. The work done by the retarding force is the net work done on the block. \[ -18 = \frac{1}{2} (1) v_f^2 - \frac{1}{2} (1) (10)^2 \] \[ -18 = \frac{1}{2} v_f^2 - \frac{1}{2} (100) \] \[ -18 = \frac{1}{2} v_f^2 - 50 \] \[ \frac{1}{2} v_f^2 = 50 - 18 = 32 \] \[ v_f^2 = 64 \] \[ v_f = \sqrt{64} = 8 \, \text{m/s} \] The final speed of the block as it crosses the rough region is 8 m/s.