This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
 Spring '08
 Turner
 Force, Friction

Click to edit the document details