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A block of mass 10 kg is moving in x-dir...

A block of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retarding force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy will be .

A

250 J

B

275 J

C

450 J

D

475 J

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The correct Answer is:
To solve the problem step by step, we will follow these steps: ### Step 1: Identify the given values - Mass of the block, \( m = 10 \, \text{kg} \) - Initial speed of the block, \( v = 10 \, \text{m/s} \) - Retarding force, \( F = -0.1x \, \text{J/m} \) - Initial position, \( x_i = 20 \, \text{m} \) - Final position, \( x_f = 30 \, \text{m} \) ### Step 2: Calculate the initial kinetic energy The initial kinetic energy (\( KE_i \)) can be calculated using the formula: \[ KE_i = \frac{1}{2} mv^2 \] Substituting the values: \[ KE_i = \frac{1}{2} \times 10 \, \text{kg} \times (10 \, \text{m/s})^2 = \frac{1}{2} \times 10 \times 100 = 500 \, \text{J} \] ### Step 3: Calculate the work done by the retarding force The work done (\( W \)) by the force as the block moves from \( x = 20 \, \text{m} \) to \( x = 30 \, \text{m} \) can be calculated using the integral of the force over the distance: \[ W = \int_{x_i}^{x_f} F \, dx \] Substituting the expression for \( F \): \[ W = \int_{20}^{30} (-0.1x) \, dx \] Calculating the integral: \[ W = -0.1 \int_{20}^{30} x \, dx = -0.1 \left[ \frac{x^2}{2} \right]_{20}^{30} \] Calculating the limits: \[ = -0.1 \left( \frac{30^2}{2} - \frac{20^2}{2} \right) = -0.1 \left( \frac{900}{2} - \frac{400}{2} \right) = -0.1 \left( 450 - 200 \right) = -0.1 \times 250 = -25 \, \text{J} \] ### Step 4: Calculate the final kinetic energy Using the work-energy principle, the final kinetic energy (\( KE_f \)) can be calculated as: \[ KE_f = KE_i + W \] Substituting the values: \[ KE_f = 500 \, \text{J} + (-25 \, \text{J}) = 500 \, \text{J} - 25 \, \text{J} = 475 \, \text{J} \] ### Final Answer The final kinetic energy of the block when it travels from \( x = 20 \, \text{m} \) to \( x = 30 \, \text{m} \) is \( 475 \, \text{J} \). ---

To solve the problem step by step, we will follow these steps: ### Step 1: Identify the given values - Mass of the block, \( m = 10 \, \text{kg} \) - Initial speed of the block, \( v = 10 \, \text{m/s} \) - Retarding force, \( F = -0.1x \, \text{J/m} \) - Initial position, \( x_i = 20 \, \text{m} \) - Final position, \( x_f = 30 \, \text{m} \) ...
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