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Homework 5 Solutions

# Homework 5 Solutions - homework 05 ALIBHAI ZAHID Due 4:00...

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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 0 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 0 θ How far does the box slide? 1. x = v 2 0 2 g ( μ cos θ + sin θ ) 2. x = v 2 0 g (sin θ μ cos θ ) 3. x = v 2 0 2 g ( μ sin θ + cos θ ) 4. x = v 2 0 g ( μ sin θ 2 cos θ ) 5. x = v 2 0 2 g sin θ 6. x = v 2 0 2 g (sin θ μ cos θ ) 7. x = v 2 0 2 g μ cos θ 8. x = v 2 0 g ( μ sin θ + cos θ ) 9. x = v 2 0 2 g ( μ sin θ cos θ ) 10. x = v 2 0 2 g ( μ cos θ sin θ ) 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 m g sin θ F f where F f is the friction force. Thus, Newton’s equation for the block reads m a = m g sin θ F f = m g sin θ μ N = m g (sin θ μ cos θ ) a = g (sin θ μ cos θ ) . where N = m g cos θ . To find the distance the block slides down the incline, use v 2 = v 2 0 + 2 a ( x x 0 ) , valid for a body moving with a constant accel- eration. Since x 0 = 0 and v f = 0 (the block stops), we get x = v 2 0 2 a = v 2 0 2 g (sin θ μ cos θ ) = v 2 0 2 g ( μ cos θ sin θ ) . Question 2 part 2 of 4 10 points How much work is done by friction? 1. W = μ m v 2 0 cos θ 2 ( μ sin θ cos θ ) 2. W = μ m g x sin θ 3. W = μ m v 0 2 ( μ tan θ ) 4. W = μ m g x tan θ 5. W = m g ( μ cos θ + sin θ ) 6. W = μ m g x cos θ correct 7. W = μ m v 0 2 ( μ + tan θ ) 8. W = m g ( μ cos θ sin θ ) 9. W = 0

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homework 05 – ALIBHAI, ZAHID – Due: Feb 21 2007, 4:00 am 2 10. W = μ m v 2 0 2 ( μ + tan θ ) Explanation: The work done by friction is W = vector f · vectorx = μ N x = μ m g x cos θ . Question 3 part 3 of 4 10 points How much work is done by the normal force? 1. W = μ m v 0 2 ( μ + tan θ ) 2. W = μ m g x cos θ 3. W = m g ( μ cos θ + sin θ ) 4. W = 0 correct 5. W = μ m v 2 0 2 ( μ + tan θ ) 6. W = μ m g x sin θ 7. W = μ m g x tan θ 8. W = μ m v 2 0 cos θ 2 ( μ sin θ cos θ ) 9. W = μ m v 0 2 ( μ tan θ ) 10. W = m g ( μ cos θ sin 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 part 4 of 4 10 points How much work is done by gravity? 1. W = m v 0 tan θ 2. W = m v 0 2 tan θ 3. W = m g x cos θ 4. W = m g (cos θ + sin θ ) 5. W = m v 2 0 tan θ 6. W = m g (cos θ sin θ ) 7. W = m g x tan θ 8. W = m v 2 0 2 tan θ 9. W = m g x sin θ correct 10. W = 0 Explanation: a = g (sin θ μ cos θ ) and x = v 2 0 2 a , so the work done by gravity is W = vector F g · vectorx = F g bardbl x = m g x sin θ . Question 5 part 1 of 3 10 points As shown in the figure, a block of mass 2 . 2 kg is pushed up against the vertical wall by a force of 64 N acting at 48 to the ceiling.
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