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Unformatted text preview: homework 15 – YOO, HEE – 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 + mg cos θ ) D 2. W = μ k N D correct 3. W = μ k ( N  mg cos θ ) D 4. W = 0 5. W = + μ k ( N  mg cos θ ) D 6. W = μ k ( N + mg cos θ ) D 7. W = + μ k N 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 D 2. W = N D 3. W = 0 correct 4. W = ( mg cos θ + F sin θN ) D 5. W = N D cos θ 6. W = ( N  mg cos θ F sin θ ) D 7. W = N D sin θ 8. W = ( N + mg cos θ + F sin θ ) D 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 μ k N ) D 3. v = radicalbigg 2 m ( F sin θ μ k N ) D 4. v = radicalbigg 2 m ( F cos θ + mg sin θ μ k N ) D 5. v = radicalbigg 2 m ( F cos θ mg sin θ μ k N ) D correct 6. v = radicalbigg 2 m ( F cos θ mg sin θ ) D 7. v = radicalbigg 2 m ( F cos θ + mg sin θ ) D 8. v = radicalbigg 2 m ( F cos θ mg sin θ + μ k N ) D Explanation: The work done by gravity is W grav = mg D cos(90 ◦ + θ ) = mg D sin θ . homework 15 – YOO, HEE – Due: Feb 23 2008, 4:00 am 2 The work done by the force F is W F = F D cos θ . From the workenergy theorem we know that W net = Δ K , W F + W grav + W friction = 1 2 mv 2 f ....
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 Spring '08
 Turner
 Energy, Force, Kinetic Energy, Mass, Work, Cos, KF

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