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Physics · WORK, ENERGY AND POWER
हिन्दीA block of mass \( m = 1\text{ kg} \) moving with initial speed \( v_i = 2\text{ m s}^{-1} \) enters a rough patch from \( x = 0.1\text{ m} \) to \( x = 2.01\text{ m} \) where it experiences a retarding force \( F_r = -\frac{k}{x} \) with \( k = 0.5\text{ J} \). Taking \( \ln(20.1) \approx 3.0 \), the speed of the block as it crosses this rough patch is:
Correct answer · Option 1
\( 1.0\text{ m s}^{-1} \)
Explanation
\( K_f = K_i + \int_{0.1}^{2.01} (-\frac{k}{x}) dx = \frac{1}{2}(1)(2)^2 - 0.5 \ln(\frac{2.01}{0.1}) = 2.0 - 0.5(3.0) = 0.5\text{ J} \). Then \( v_f = \sqrt{2K_f/m} = \sqrt{2(0.5)/1} = 1.0\text{ m s}^{-1} \).
