Kaplunovsky - practice 07

Kaplunovsky - practice 07 - practice 07 Due: Mar 5 2007,...

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Unformatted text preview: practice 07 Due: Mar 5 2007, noon 1 Question 1, chap -1, sect -1. part 1 of 5 5 points A 8 . 37 kg block is pushed 9 . 4 m up a vertical wall with constant speed by a constant force F applied at an angle of 58 with the horizontal. 8 . 37 kg F 5 8 The coefficient of kinetic friction between the wall and the block is = 0 . 38; accelera- tion of gravity is g = 9 . 8 m / s 2 . Calculate the magnitude of the force F . Correct answer: 126 . 842 N (tolerance 1 %). Explanation: Given : m = 8 . 37 kg , = 0 . 38 , = 58 , and y = 9 . 4 m . F 5 8 v mg f k n The block is in equilibrium horizontally, hence summationdisplay F x = F cos n = 0 , and therefore n = F cos . Consequently, the friction force f k = n = F cos . Vertically, the block moves with constant ve- locity, v y = const = a y = 0, hence summationdisplay F y = F sin mg f k = 0 , F sin mg F cos = 0 , F (sin cos ) = mg, and therefore, F = mg sin cos = (8 . 37 kg) (9 . 8 m / s 2 ) sin58 . 38 cos58 = 126 . 842 N Question 2, chap -1, sect -1. part 2 of 5 5 points Find the work done by the force F . Correct answer: 1011 . 14 J (tolerance 1 %). Explanation: THe work of the force vector F is W = vector F vector R = F y Y where the second equality follows from the displacement vector R being in the y direction only. Since F y = F sin , we have W = ( F sin ) Y = (126 . 842 N) sin(58 ) (9 . 4 m) = 1011 . 14 J . Question 3, chap -1, sect -1. part 3 of 5 5 points Find the work done by the force of gravity. Correct answer: 771 . 044 J (tolerance 1 %). Explanation: The force of gravity has opposite direction to the displacement, hence its work is W g = mvectorg vector R = mg Y cos(180 ) = mg Y = (8 . 37 kg)(9 . 8 m / s 2 )(9 . 4 m) = 771 . 044 J . practice 07 Due: Mar 5 2007, noon 2 Question 4, chap -1, sect -1. part 4 of 5 5 points Find the work done by the normal force between the block and the wall. Correct answer: 0 J (tolerance 1 %). Explanation: The normal force is perpendicular to the displacement, so its work is zero: W n = vectorn vector R = n Y cos(90 ) = 0 . Question 5, chap -1, sect -1. part 5 of 5 5 points By how much does the gravitational poten- tial energy increase during this motion? Correct answer: 771 . 044 J (tolerance 1 %). Explanation: U g = mg y = (8 . 37 kg) (9 . 8 m / s 2 ) (9 . 4 m) = 771 . 044 J . Question 6, chap 8, sect 5. part 1 of 2 5 points One of the most powerful cranes in the world, operating in Switzerland, can slowly raise a load of 6000 tonne to a height of 10 m....
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Kaplunovsky - practice 07 - practice 07 Due: Mar 5 2007,...

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