Question

A block of mass 2 kg moving on a horizontal surface with speed of 4 ms^–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 Nm^–1. The speed of the block as it just crosses the rough surface will be

• A. Zero
• B. 1.5 ms^–1
• C. 2.0 ms^–1
• D. 2.5 ms^–1

Answer 1

The Correct Option is C

To determine the speed of the block as it just crosses the rough surface, we need to analyze the forces and energy changes it experiences. The block initially moves at a speed of 4 m/s and encounters a retarding force over a specific range of distance.

Step 1: Initial Conditions

• Mass of the block, m = 2 \, \text{kg}
• Initial speed, v_i = 4 \, \text{m/s}
• The retarding force is given by F = -kx, where k = 12 \, \text{N/m}
• The rough surface extends from x = 0.5 \, \text{m} to x = 1.5 \, \text{m}

Step 2: Work Done by the Retarding Force

• The force F = -kx does work over the distance the block travels.
• The work done by the force is calculated using the integral:

\text{Work Done} = \int_{0.5}^{1.5} (-kx) \, dx

• Substitute k = 12 \, \text{N/m}:

\text{Work Done} = \int_{0.5}^{1.5} (-12x) \, dx = -12 \left[ \frac{x^2}{2} \right]_{0.5}^{1.5}

= -12 \left( \frac{(1.5)^2}{2} - \frac{(0.5)^2}{2} \right)

= -12 \left( \frac{2.25}{2} - \frac{0.25}{2} \right)

= -12 \times \left( 1 - 0.125 \right) = -12 \times 0.875 = -10.5 \, \text{Joules}

Step 3: Application of Work-Energy Principle

• The initial kinetic energy of the block is given by:

KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 2 \times (4)^2 = 16 \, \text{Joules}

• The work-energy principle states that the work done by the forces is equal to the change in kinetic energy:

\text{Work Done} = KE_f - KE_i

-10.5 = KE_f - 16

• Solve for KE_f (final kinetic energy):

KE_f = 16 - 10.5 = 5.5 \, \text{Joules}

Step 4: Calculate the Final Speed

• Using the final kinetic energy:

KE_f = \frac{1}{2} m v_f^2

• Substitute KE_f = 5.5 \, \text{Joules} and m = 2 \, \text{kg}:

5.5 = \frac{1}{2} \times 2 \times v_f^2

5.5 = v_f^2

• Solving for v_f gives:

v_f = \sqrt{5.5} \approx 2.35 \, \text{m/s}

Therefore, rounding to the nearest option, the speed of the block as it just crosses the rough surface is approximately 2.0 m/s.