This preview shows pages 1–2. 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: schneiter (tcs639) HW08 Ross (89152) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A block accelerates 3 . 5 m / s 2 down a plane inclined at angle 26.0 . The acceleration of gravity is 9 . 81 m / s 2 . m k 3 . 5 m / s 2 26 Find k between the block and the inclined plane. Correct answer: 0 . 0907798. Explanation: Given : a = 3 . 5 m / s 2 , = 26 , and g = 9 . 81 m / s 2 . Consider the free body diagram for the block m g s i n N = m g c o s N a m g Basic Concepts: vector F net = mvectora Parallel to the ramp: F x,net = ma x = F g,x- F k F g,x = mg sin F k = k F n Perpendicular to the ramp: F y,net = F n- F g,y = 0 F g,y = mg cos Solution: Consider the forces parallel to the ramp: F k = F g,x- ma x = mg sin - ma x Consider the forces perpendicular to the ramp: F n = F g,y = mg cos Thus the coefficient of friction is k = F k F n = m g sin - m a m g cos = g sin - a g cos = 9 . 81 m / s 2 sin 26 - 3 . 5 m / s 2 9 . 81 m / s 2 cos 26 = . 0907798 . 002 10.0 points The magnitude of each force is 283 N, the force on the right is applied at an angle 35 and the mass of the block is 23 kg. The coefficient of friction is 0 . 148....
View Full Document
This note was uploaded on 04/23/2009 for the course PHYS 152 taught by Professor Button during the Spring '08 term at IUPUI.
- Spring '08