hw5sol - homework 05 ARMSTRONG, DOMINIC Due: Oct 9 2007,...

Info iconThis preview shows pages 1–3. 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: homework 05 ARMSTRONG, DOMINIC Due: Oct 9 2007, 3:00 am 1 Question 1, chap 7, sect 1. part 1 of 4 10 points A box of mass m with an initial velocity of v slides down a plane, inclined at with respect to the horizontal. The coefficient of kinetic friction is . The box stops after sliding a distance x . m k v How far does the box slide? 1. x = v 2 g ( sin - 2 cos ) 2. x = v 2 2 g ( sin + cos ) 3. x = v 2 2 g sin 4. x = v 2 2 g (sin - cos ) 5. x = v 2 2 g ( cos + sin ) 6. x = v 2 g (sin - cos ) 7. x = v 2 g ( sin + cos ) 8. x = v 2 2 g ( cos - sin ) correct 9. x = v 2 2 g cos 10. x = v 2 2 g ( sin - cos ) Explanation: Basic Concepts: Motion under constant force W = vector F vectors Solution: The net force on the block parallel to the incline is F net = F mg sin - F f where F f is the friction force. Thus, Newtons equation for the block reads ma = mg sin - F f = mg sin - N = mg (sin - cos ) a = g (sin - cos ) . where N = mg cos . To find the distance the block slides down the incline, use v 2 = v 2 + 2 a ( x- x ) , valid for a body moving with a constant accel- eration. Since x = 0 and v f = 0 (the block stops), we get x =- v 2 2 a =- v 2 2 g (sin - cos ) = v 2 2 g ( cos - sin ) . Question 2, chap 7, sect 1. part 2 of 4 10 points How much work is done by friction? 1. W =- mg x tan 2. W =- mv 2 ( + tan ) 3. W = 0 4. W =- mv 2 ( - tan ) 5. W =- mg x sin 6. W =- mg ( cos + sin ) 7. W =- mg x cos correct 8. W =- mv 2 2 ( + tan ) 9. W =- mv 2 cos 2 ( sin - cos ) homework 05 ARMSTRONG, DOMINIC Due: Oct 9 2007, 3:00 am 2 10. W =- mg ( cos - sin ) Explanation: The work done by friction is W = vector f vectorx =- N x =- mg x cos . Question 3, chap 7, sect 1. part 3 of 4 10 points How much work is done by the normal force? 1. W =- mg ( cos - sin ) 2. W =- mv 2 ( + tan ) 3. W = 0 correct 4. W =- mg ( cos + sin ) 5. W =- mg x tan 6. W =- mv 2 ( - tan ) 7. W =- mv 2 2 ( + tan ) 8. W =- mg x cos 9. W =- mg x sin 10. W =- mv 2 cos 2 ( sin - cos ) Explanation: The normal force N is always perpendicular to the motion, so the work done by N is W = vector N vectorx = 0 . Question 4, chap 7, sect 1. part 4 of 4 10 points How much work is done by gravity? 1. W = mg x sin correct 2. W = mv 2 2 tan 3. W = mg (cos + sin ) 4. W = mg (cos - sin ) 5. W = mv 2 tan 6. W = mv 2 tan 7. W = mg x tan 8. W = mv tan 9. W = 0 10. W = mg x cos Explanation: a = g (sin - cos ) and x =- v 2 2 a , so the work done by gravity is W = vector F g vectorx = F g bardbl x = mg x sin ....
View Full Document

Page1 / 7

hw5sol - homework 05 ARMSTRONG, DOMINIC Due: Oct 9 2007,...

This preview shows document pages 1 - 3. Sign up to view the full document.

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