solution-hw8

solution-hw8 - lai (yl8859) HW08 Ross (15200) 1 This...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: lai (yl8859) HW08 Ross (15200) 1 This print-out should have 5 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The magnitude of each force is 240 N, the force on the right is applied at an angle 35 and the mass of the block is 20 kg. The coefficient of friction is 0 . 276. The acceleration of gravity is 9 . 8 m / s 2 . 20 kg = 0 . 276 2 4 N 35 240 N What is the magnitude of the resulting ac- celeration? Correct answer: 21 . 0247 m / s 2 . Explanation: Given : F = 240 N , = 35 , = 0 . 276 , m = 20 kg , and g = 9 . 8 m / s 2 , m F N m g F N The block is in equilibrium vertically, so summationdisplay F up = summationdisplay F down F sin + N = m g N = m g- F sin The block is accelerating horizontally, so F net = m a = F + F cos - N F net = m a = F (1 + cos + sin )- m g a = F m parenleftBig 1 + cos + sin parenrightBig- g = 240 N 20 kg bracketleftBig 1 + cos 35 + (0 . 276) sin35 bracketrightBig- (0 . 276) (9 . 8 m / s 2 ) = 21 . 0247 m / s 2 . 002 (part 1 of 2) 10.0 points Two blocks are attached by a thin inextensi- ble string over a frictionless, massless pulley....
View Full Document

This note was uploaded on 02/24/2011 for the course PHYS 152 taught by Professor Button during the Winter '08 term at IUPUI.

Page1 / 3

solution-hw8 - lai (yl8859) HW08 Ross (15200) 1 This...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online