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11.A man is pulling a block of 10kg from rest with 80N up an inclined slope.The length of the slope is 10m and the friction acting on the block is 10N.The slope has 30 degree.(i)How much time does the man take to pull the block to the top?(ii)What is the average power of the man?(iii)Find the gain in the... 顯示更多 11.A man is pulling a block of 10kg from rest with 80N up an inclined slope.The length of the slope is 10m and the friction acting on the block is 10N.The slope has 30 degree. (i)How much time does the man take to pull the block to the top? (ii)What is the average power of the man? (iii)Find the gain in the kinetic energy of the block when it reaches the top of the slope. (iv)After reaching the top of the slope, he pulls the block on the plane surface with constant power which is equal to that calculated in (ii).Find the terminal speed of the block.

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最佳解答:

(i) Net force acting on the block = ( 80 - 10 - 10 g.sin(30)) N = 20 N where g is the acceleration due to gravity, taken to be 10 m /s2 Acceleration = 20/10 ms/2 = 2 m /s2 Use equation of motion: s = ut + (1/2).a.t^2 with u = 0 m /2, a = 2 m /s2, s = 10 m , t = ? hence, 10 = (1/2).(2).t^2 i.e. t = 3.162 s (ii) Work done by the man = 80 x 10 J = 800 J Average power = work done/time = 800/3.162 watts = 253 watts (iii) Use equation of motion: v^2 = u^2 + 2.a.s with u = 0 m /2, a = 2 m /s2, s = 10 m , v = ? hence, v^2 = 2 x 2 x 10 (m/s)^2 i.e. v = 6.325 m /s Gain in kinetic energy = (1/2) x 10 x 40 J = 200 J (iv) Let v be the terminal speed, use power = froce x velocity assume the frictional force remains the same,i.e. 10 N hence, 10v = 253 2010-05-15 23:58:39 補充： i.e. v = 253/10 m/s 25.3 m/s

其他解答:

Answer: (i)3.16s (ii)253W (iii)200J (iv)25.3ms^-1|||||i. Along the plane, net force acting on the block, F = 80 - 10(10)sin30* - 10 = 20 N By Newton's 2nd law of motion, F = ma 20 = 10a Acceleration, a = 2 ms^-2 By s = ut + 1/2 at^2 10 = 0 + 1/2 (2)t^2 Time, t = 3.16 s ii. Gain in G.P.E. = mgh = (10)(10)(10sin30*) = 500 J Average power, P = E/t = 500/3.16 = 158 W iii. By v^2 = u^2 + 2as v^2 = 0 + 2(2)(10) v = 6.32 ms^-1 Gain in K.E. = 1/2 mv^2 = 1/2 (10)(6.32)^2 = 200 J iv. Assume the man exerts the same force on the block By P = Fv 158 = 80v Terminal speed, v = 1.98 ms^-17638E748CCC65837

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