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想問下呢幾條數點做,請例步驟!?如果唔例我會唔明
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Question 1A 4.0 kg book is placed on a long straight plank. One end of the plank is slowly raised, to the point where the book just begins to slip. The angle at that point is 170. (a) Make a sketch of the free body diagram to show all forces acting on the book.(b) Calculate the coefficient of static... 顯示更多 Question 1 A 4.0 kg book is placed on a long straight plank. One end of the plank is slowly raised, to the point where the book just begins to slip. The angle at that point is 170. (a) Make a sketch of the free body diagram to show all forces acting on the book. (b) Calculate the coefficient of static friction. (c) The book is observed to accelerate down the plank, reaching a speed of 1.5 m/s after slipping a distance of D = 2.5 m (measured parallel the plank). Calculate the total work done on the book over that distance. (d) Calculate the work done by gravity over the distance D. (e) Is there any work done by the normal force? Why? Question 2 The same constant force F is used to accelerate two carts of the same mass at rest on frictionless tracks. If distance of the force applied to cart A is twice as long as that applied to cart B. Calculate the ratio of their final velocities.
最佳解答:
Q1 (a) There are three forces acting on the block. -- the weight of the block, vertically downward -- the normal reaction, upward perpenduclar to the plank surface -- friction, upward along the plank surface (b) Consider direction along the plank surface, since the block is at rest, upward force = downward force firction Ff = 4g.sin(17) where g is the acceleration due to gravity But Ff = uR, where u is the coefficient of static friction and R is the normal reaction hence, u = 4g.sin(17)/R = 4g.sin(17)/[4g.cos(17)] = tan(17) = 0.306 (c) work done = increase in kinetic energy (KE) of the block i.e. work done = (1/2)x(4)x(1.5)^2 J = 4.5 J (d) The downward fall of the block = 2.5.sin(17) m thus, work done by gravity = 4g x (2.5.sin(17)) J = 29.2 J (e) No. The normal reaction is always perpendicular to the direction of travel of the block. There is no displacement of the normal reaction along its direction of action, hence no work is done. ------------------------------------------------------------- Q2 Let m be the mass of each car d be the distance where the force applied to car B thus distance of force application to car A = 2d using work-done = change of kinetic energy For car A: F(2d) = (1/2)mVa^2 ---------------- (1) For car B: F(d) = (1/2)mVb^2 ---------------- (2) where Va and Vb are the final velocities of cars A and B respectively (1)/(2): 2 = [Va/Vb]^2 thus, Va/Vb = sqaure-root(2)
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