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Step 1: Identify the Given Data
• Mass of the block, $m = 10\text{ kg}$
• Initial speed, $v_i = 10\text{ m/s}$
• Retarding force, $F = 0.1x \text{ J/m}$ (acting in the opposite direction of motion)
• Initial position, $x_1 = 20\text{ m}$
• Final position, $x_2 = 30\text{ m}$
Step 2: Calculate the Initial Kinetic Energy
The initial kinetic energy $K_i$ is given by
$$K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 10 \times (10)^2 = 500\text{ J}.$$
Step 3: Compute the Work Done by the Retarding Force
The work done $W$ by a force $F(x)$ from $x = x_1$ to $x = x_2$ is
$$W = \int_{x_1}^{x_2} F(x)\,dx.$$
Since the force is retarding (opposite to the direction of motion), we take $F_r = -0.1x$. Thus,
$$W = \int_{20}^{30} -0.1x \, dx = -0.1 \int_{20}^{30} x \, dx.$$
Evaluate this integral:
$$\int_{20}^{30} x \, dx = \left[ \frac{x^2}{2} \right]_{20}^{30} = \frac{30^2}{2} - \frac{20^2}{2} = \frac{900}{2} - \frac{400}{2} = 450 - 200 = 250.$$
Therefore,
$$W = -0.1 \times 250 = -25\text{ J}.$$
Step 4: Apply the Work-Energy Theorem
The work-energy theorem states:
$$W = K_f - K_i.$$
Here $K_f$ is the final kinetic energy. So,
$$K_f = K_i + W = 500\text{ J} + (-25\text{ J}) = 475\text{ J}.$$
Final Answer
The final kinetic energy of the block is $475\text{ J}$.
