OnlineCBT
Physics · LAWS OF MOTION
हिन्दीA block of mass \( m = 4\text{ kg} \) is projected with an initial velocity \( v_0 = 10\text{ m/s} \) along a rough horizontal plane. The block comes to rest after covering a distance of \( d = 25\text{ m} \). Taking \( g = 10\text{ m/s}^2 \), calculate the coefficient of kinetic friction \( \mu_k \) and the rate of work done by friction (power) at time \( t = 1\text{ s} \).
Correct answer · Option 4
\( \mu_k = 0.2 \) and \( P = 64\text{ W} \)
Explanation
Deceleration \( a = \frac{v_0^2}{2d} = \frac{100}{50} = 2\text{ m/s}^2 \implies \mu_k = \frac{a}{g} = \frac{2}{10} = 0.2 \). At \( t = 1\text{ s} \), velocity \( v = 10 - 2(1) = 8\text{ m/s} \). Friction force \( f_k = \mu_k m g = 0.2 \times 4 \times 10 = 8\text{ N} \). Instantaneous power magnitude \( P = f_k v = 8 \times 8 = 64\text{ W} \).
