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An 8.5-kg block is pushed along a horizontal rough surface by a 40-N force inclined at 20° with the horizontal. The coefficient of friction between the surface and block is 0.35. If the block has an initial velocity of 3.6 m/s and the force does 200 J of work on the block, find: (a) The total distance moved by the block. (b) The final velocity of the block. |
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Answer» Mass of the block m = 8.5kgForce F = 40N.θ = 20°Coefficient of friction μ =0.35Initial Velocity of block v = 3.6m/s.WORK DONE by force F, W = 200J.To Find:-a) The total distance moved by the blockb) The final velocity of the blockSolution:-a) Work Done by the force W = Fdcosθ200 = 40xdcos20°By solve we get d = 5.32mb) From the free body diagram ( Find attached FIGURE) = Friction Force = μNthen , N = mg + FsinθN = 8.5x9.81 + θ0sin20°N = 97.06N = 0.35x97.06N = 33.973N = 40cos20° - 33.973 = 3.6146As per Newton's law = m x acceleration(a)a = a = 0.4252m/s²We have v² = u² + 2asv² = (3.6)² + 2(0.4252)(5.32) {s= d = 5.32}v² = 12.96 + 4.524128v² = 17.4841v = 4.181m/sec |
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