HW15 - homework 15 PAPAGEORGE MATT Due 4:00 am Question 1...

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homework 15 – PAPAGEORGE, MATT – Due: Feb 23 2008, 4:00 am 1 Question 1, chap 7, sect 3. part 1 of 3 10 points A block of mass m is pushed a distance D up an inclined plane by a horizontal force F . The plane is inclined at an angle θ with respect to the horizontal. The block starts from rest and the coefficient of kinetic friction is μ k . m D μ k F θ If N is the normal force, what is the work done by friction? 1. W = + μ k N D 2. W = + μ k ( N - m g cos θ ) D 3. W = 0 4. W = - μ k ( N + m g cos θ ) D 5. W = - μ k N D correct 6. W = + μ k ( N + m g cos θ ) D 7. W = - μ k ( N - m g cos θ ) D Explanation: The force of friction has a magnitude F friction = μ k N . Since it is in the direc- tion opposite to the motion, we get W friction = - F friction D = - μ k N D. Question 2, chap 7, sect 3. part 2 of 3 10 points What is the work done by the normal force N ? 1. W = ( N + m g cos θ + F sin θ ) D 2. W = -N D 3. W = 0 correct 4. W = N D 5. W = N D sin θ 6. W = ( N - m g cos θ - F sin θ ) D 7. W = ( m g cos θ + F sin θ - N ) D 8. W = N D cos θ Explanation: The normal force makes an angle of 90 with the displacement, so the work done by it is zero. Question 3, chap 7, sect 3. part 3 of 3 10 points What is the final speed of the block? 1. v = radicalbigg 2 m ( F sin θ + μ k N ) D 2. v = radicalbigg 2 m ( F cos θ - m g sin θ ) D 3. v = radicalbigg 2 m ( F cos θ - m g sin θ - μ k N ) D correct 4. v = radicalbigg 2 m ( F cos θ - m g sin θ + μ k N ) D 5. v = radicalbigg 2 m ( F sin θ - μ k N ) D 6. v = radicalbigg 2 m ( F cos θ - μ k N ) D 7. v = radicalbigg 2 m ( F cos θ + m g sin θ - μ k N ) D 8. v = radicalbigg 2 m ( F cos θ + m g sin θ ) D Explanation: The work done by gravity is W grav = m g D cos(90 + θ ) = - m g D sin θ .
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homework 15 – PAPAGEORGE, MATT – Due: Feb 23 2008, 4:00 am 2 The work done by the force F is W F = F D cos θ . From the work-energy theorem we know that W net = Δ K , W F + W grav + W friction = 1 2 m v 2 f . Thus v f = radicalbigg 2 m ( F cos θ - m g sin θ - μ k N ) D . Question 4, chap 7, sect 3. part 1 of 1 10 points A rock of mass m is thrown horizontally off a building from a height h . The speed of the rock as it leaves the thrower’s hand at the edge of the building is v 0 , as shown.
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