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http://i44.photobucket.com/albums/f27/eric19911991/-2.jpga man is pulling a block of 10kg from rest with 80N up an inclined road (fig above)the length of the road is 10 m and friction acting on the block is 10Nquestions: a)how much time does the man take to pull the block to the top ? ... 顯示更多 http://i44.photobucket.com/albums/f27/eric19911991/-2.jpg a man is pulling a block of 10kg from rest with 80N up an inclined road (fig above) the length of the road is 10 m and friction acting on the block is 10N questions: a)how much time does the man take to pull the block to the top ? b)what is the power of the man? c)fine the gain of KEof the block when it reaches the top of the road

最佳解答:

(a) First of all, we have to find out the resultant force on the block as follows: Component of its weight along the inclined plane = mg sin θ = 10 × 10 × sin 30° = 50N Friction on the block = 10N Resultant force on the block = 80 - 50 - 10 = 20 N up the inclined plane. So the acceleration of the block is: 20/10 = 2 m/s2 up the inclined plane Then, applying the equation s = ut + (1/2)at2, with u = 0, s = 10 and a = 2: 10 = (1/2) × 2 × t2 t = 3.16 s (b) In fact, since the velocity of the block is changing in the motion time, the power of the man is varying too during the motion according to the equation Power = Force × Velocity. Therefore, instead, we may find out his average power as follows: Time of action of the man's force = 3.16 s Velocity of the block when it reaches the top = 3.16 × 2 = 6.32 m/s K.E. of the block = (1/2) × 10 × 6.322 = 200 J Potential energy gain of the block = 10 × 10 × 10 sin 30° = 500 J Work done against friction = 10 × 10 = 100 J So total energy output by the man = 200 + 500 + 100 = 800 J Therefore, his average power is 800/3.16 = 253 W (c) As found in part (b), K.E. = 200 J

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