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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 ....
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 Fall '07
 Swinney
 mechanics, Mass, Work

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