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 rough region is

The motion is along the +x-direction. Outside the rough strip the block is free; inside the strip (from $$x_i = 0.1\;{\rm m}$$ to $$x_f = 1.9\;{\rm m}$$) it experiences the position-dependent retarding force

$$F_r = -\,k\,x,$$ with $$k = 10\;{\rm N\,m^{-1}}$$. (The minus sign shows that the force is opposite to the direction of motion, so it removes kinetic energy from the block.)

Case 1 : Work done by the retarding force

The infinitesimal work done by the force over a small displacement $$dx$$ is $$dW = F_r\,dx = -k\,x\,dx.$$ To obtain the total work between the entry and exit points, integrate from $$x = 0.1$$ m to $$x = 1.9$$ m:

$$\begin{aligned} W &= \int_{0.1}^{1.9} (-k\,x)\,dx \\ &= -k \int_{0.1}^{1.9} x\,dx \\ &= -k\left[\frac{x^{2}}{2}\right]_{0.1}^{1.9} \\ &= -\frac{k}{2}\Bigl(x_f^{2}-x_i^{2}\Bigr). \end{aligned}$$

Insert the numerical values:

$$\begin{aligned} x_f^{2}-x_i^{2} &= (1.9)^{2}-(0.1)^{2}=3.61-0.01=3.60,\\[4pt] W &= -\frac{10}{2}\,(3.60) = -5 \times 3.60 = -18\;{\rm J}. \end{aligned}$$

The negative sign confirms that the force takes 18 J of mechanical energy out of the block.

Case 2 : Applying the work-energy theorem

The work-energy theorem states $$W = K_f - K_i,$$ where $$K_i$$ and $$K_f$$ are the initial and final kinetic energies, respectively.

Initial kinetic energy: $$K_i = \frac{1}{2} m v_i^{2} = \tfrac12 (1\;{\rm kg})(10\;{\rm m/s})^{2} = 50\;{\rm J}.$$

Therefore,

$$\begin{aligned} -18\;{\rm J} &= K_f - 50\;{\rm J}\\ K_f &= 50\;{\rm J} - 18\;{\rm J} = 32\;{\rm J}. \end{aligned}$$

Case 3 : Determining the final speed

Let the required speed be $$v_f$$. $$K_f = \frac{1}{2} m v_f^{2} \; \Longrightarrow\; \frac{1}{2}(1)\,v_f^{2} = 32.$$

Solve for $$v_f$$:

$$v_f^{2} = 64 \quad\Longrightarrow\quad v_f = 8\;{\rm m/s}.$$

Hence, when the block leaves the rough region its speed is 8 m s−1.

Final answer : 8 m/s (Option D)