hw5sol - homework 05 – ARMSTRONG DOMINIC – Due Oct 9...

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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 θ with respect to the horizontal. The coefficient of kinetic friction is μ . The box stops after sliding a distance x . m μ k v θ How far does the box slide? 1. x = v 2 g ( μ sin θ- 2 cos θ ) 2. x = v 2 2 g ( μ sin θ + cos θ ) 3. x = v 2 2 g sin θ 4. x = v 2 2 g (sin θ- μ cos θ ) 5. x = v 2 2 g ( μ cos θ + sin θ ) 6. x = v 2 g (sin θ- μ cos θ ) 7. x = v 2 g ( μ sin θ + cos θ ) 8. x = v 2 2 g ( μ cos θ- sin θ ) correct 9. x = v 2 2 g μ cos θ 10. x = v 2 2 g ( μ sin θ- cos θ ) 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 θ- F f where F f is the friction force. Thus, Newton’s equation for the block reads ma = mg sin θ- F f = mg sin θ- μ N = mg (sin θ- μ cos θ ) a = g (sin θ- μ cos θ ) . where N = mg cos θ . 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 θ- μ cos θ ) = v 2 2 g ( μ cos θ- sin θ ) . Question 2, chap 7, sect 1. part 2 of 4 10 points How much work is done by friction? 1. W =- μmg x tan θ 2. W =- μmv 2 ( μ + tan θ ) 3. W = 0 4. W =- μmv 2 ( μ- tan θ ) 5. W =- μmg x sin θ 6. W =- mg ( μ cos θ + sin θ ) 7. W =- μmg x cos θ correct 8. W =- μmv 2 2 ( μ + tan θ ) 9. W =- μmv 2 cos θ 2 ( μ sin θ- cos θ ) homework 05 – ARMSTRONG, DOMINIC – Due: Oct 9 2007, 3:00 am 2 10. W =- mg ( μ cos θ- sin θ ) Explanation: The work done by friction is W = vector f · vectorx =- μ N x =- μmg x cos θ . Question 3, chap 7, sect 1. part 3 of 4 10 points How much work is done by the normal force? 1. W =- mg ( μ cos θ- sin θ ) 2. W =- μmv 2 ( μ + tan θ ) 3. W = 0 correct 4. W =- mg ( μ cos θ + sin θ ) 5. W =- μmg x tan θ 6. W =- μmv 2 ( μ- tan θ ) 7. W =- μmv 2 2 ( μ + tan θ ) 8. W =- μmg x cos θ 9. W =- μmg x sin θ 10. W =- μmv 2 cos θ 2 ( μ sin θ- cos θ ) 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 θ correct 2. W = mv 2 2 tan θ 3. W = mg (cos θ + sin θ ) 4. W = mg (cos θ- sin θ ) 5. W = mv 2 tan θ 6. W = mv 2 tan θ 7. W = mg x tan θ 8. W = mv tan θ 9. W = 0 10. W = mg x cos θ Explanation: a = g (sin θ- μ cos θ ) 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 θ ....
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This homework help was uploaded on 04/07/2008 for the course PHY 301 taught by Professor Swinney during the Fall '07 term at University of Texas.

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hw5sol - homework 05 – ARMSTRONG DOMINIC – Due Oct 9...

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