A block of mass 2 kg moving on a horizontal surface with speed of 4 m s$$^{-1}$$ enters a rough surface ranging from $$x = 0.5$$ m to $$x = 1.5$$ m. The retarding force in this range of rough surface is related to distance by $$F = -kx$$ where $$k = 12$$ N m$$^{-1}$$. The speed of the block as it just crosses the rough surface will be

A block of mass $$m = 2$$ kg has initial speed $$v_0 = 4$$ m s$$^{-1}$$. It encounters a rough surface from $$x = 0.5$$ m to $$x = 1.5$$ m with retarding force $$F = -kx$$, where $$k = 12$$ N m$$^{-1}$$. The initial kinetic energy is $$KE_i = \frac{1}{2}mv_0^2 = \frac{1}{2}(2)(4^2) = 16 \text{ J}$$.

The work done by the retarding force is $$W = \int_{0.5}^{1.5} F \, dx = -\int_{0.5}^{1.5} 12x \, dx = -12 \left[\frac{x^2}{2}\right]_{0.5}^{1.5}$$ which gives $$W = -6\left[(1.5)^2 - (0.5)^2\right] = -6[2.25 - 0.25] = -6 \times 2 = -12 \text{ J}$$. Applying the work-energy theorem, $$KE_f = KE_i + W = 16 - 12 = 4 \text{ J}$$ so that $$\frac{1}{2}mv^2 = 4$$ and hence $$v = \sqrt{\frac{2 \times 4}{2}} = \sqrt{4} = 2 \text{ m s}^{-1}$$. The correct answer is Option A.