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Physics · WORK, ENERGY AND POWER
हिन्दीA block of mass \( m = 2\text{ kg} \) moving with speed \( v_i = 4\text{ m s}^{-1} \) enters a rough patch from \( x = 0.5\text{ m} \) to \( x = 5.0\text{ m} \). The retarding force acting on the block in this region is \( F(x) = -\frac{k}{x} \), where \( k = 4.0\text{ J} \). (Given: \( \ln 10 = 2.30 \)). The speed of the block as it emerges from the patch at \( x = 5.0\text{ m} \) is:
Correct answer · Option 2
\( \sqrt{6.8}\text{ m s}^{-1} \)
Explanation
\( K_f = K_i + \int_{0.5}^5 (-\frac{k}{x})dx = \frac{1}{2}(2)(4)^2 - k \ln(\frac{5.0}{0.5}) = 16 - 4.0(2.30) = 16 - 9.2 = 6.8\text{ J} \). Then \( v_f = \sqrt{2K_f/m} = \sqrt{2(6.8)/2} = \sqrt{6.8}\text{ m s}^{-1} \).
