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Unformatted text preview: homework 05 ALIBHAI, ZAHID Due: Feb 21 2007, 4:00 am 1 Question 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 2 g ( cos + sin ) 2. x = v 2 g (sin cos ) 3. x = v 2 2 g ( sin + cos ) 4. x = v 2 g ( sin 2 cos ) 5. x = v 2 2 g sin 6. x = v 2 2 g (sin cos ) 7. x = v 2 2 g cos 8. x = v 2 g ( sin + cos ) 9. x = v 2 2 g ( sin cos ) 10. x = v 2 2 g ( cos sin ) correct 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 part 2 of 4 10 points How much work is done by friction? 1. W = mv 2 cos 2 ( sin cos ) 2. W = mg x sin 3. W = mv 2 ( tan ) 4. W = mg x tan 5. W = mg ( cos + sin ) 6. W = mg x cos correct 7. W = mv 2 ( + tan ) 8. W = mg ( cos sin ) 9. W = 0 homework 05 ALIBHAI, ZAHID Due: Feb 21 2007, 4:00 am 2 10. W = mv 2 2 ( + tan ) Explanation: The work done by friction is W = vector f vectorx = N x = mg x cos . Question 3 part 3 of 4 10 points How much work is done by the normal force? 1. W = mv 2 ( + tan ) 2. W = mg x cos 3. W = mg ( cos + sin ) 4. W = 0 correct 5. W = mv 2 2 ( + tan ) 6. W = mg x sin 7. W = mg x tan 8. W = mv 2 cos 2 ( sin cos ) 9. W = mv 2 ( tan ) 10. W = mg ( cos sin ) 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 part 4 of 4 10 points How much work is done by gravity? 1. W = mv tan 2. W = mv 2 tan 3. W = mg x cos 4. W = mg (cos + sin ) 5. W = mv 2 tan 6. W = mg (cos sin ) 7. W = mg x tan 8. W = mv 2 2 tan 9. W = mg x sin correct 10. W = 0 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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 Spring '08
 Turner
 Friction, Mass, Work

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