HW05-solutions - dang (ttd73) HW05 Tsoi (58020) 1 This...

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Unformatted text preview: dang (ttd73) HW05 Tsoi (58020) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 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 with respect to the horizontal. The block starts from rest and the coefficient of kinetic friction is k . m D k F If N is the normal force, what is the work done by friction? 1. W = + k N D 2. W =- k ( N - mg cos ) D 3. W = + k ( N - mg cos ) D 4. W = + k ( N + mg cos ) D 5. W =- k ( N + mg cos ) D 6. W =- k N D correct 7. W = 0 Explanation: The force of friction has a magnitude F friction = k N . Since it is in the direc- tion opposite to the motion, we get W friction =- F friction D =- k N D. 002 (part 2 of 3) 10.0 points What is the work done by the normal force N ? 1. W = ( N - mg cos - F sin ) D 2. W = ( mg cos + F sin - N ) D 3. W = N D cos 4. W = N D sin 5. W =-N D 6. W = 0 correct 7. W = N D 8. W = ( N + mg cos + F sin ) D Explanation: The normal force makes an angle of 90 with the displacement, so the work done by it is zero. 003 (part 3 of 3) 10.0 points What is the final speed of the block? 1. v = radicalbigg 2 m ( F cos + mg sin ) D 2. v = radicalbigg 2 m ( F sin + k N ) D 3. v = radicalbigg 2 m ( F sin - k N ) D 4. v = radicalbigg 2 m ( F cos - mg sin + k N ) D 5. v = radicalbigg 2 m ( F cos - mg sin - k N ) D correct 6. v = radicalbigg 2 m ( F cos - k N ) D 7. v = radicalbigg 2 m ( F cos - mg sin ) D 8. v = radicalbigg 2 m ( F cos + mg sin - k N ) D Explanation: The work done by gravity is W grav = mg D cos(90 + ) =- mg D sin . The work done by the force F is W F = F D cos . dang (ttd73) HW05 Tsoi (58020) 2 From the work-energy theorem we know that W net = K , W F + W grav + W friction = 1 2 mv 2 f . Thus v f = radicalbigg 2 m ( F cos - mg sin - k N ) D. 004 (part 1 of 2) 10.0 points A 500-N crate needs to be lifted 1 meter ver- tically in order to get it into the back of a pickup truck. What gives the crate a greater potential energy? 1. Either correct 2. Unable to determine 3. lift it straight up into the truck 4. slide it up a frictionless inclined plane Explanation: The change in height is the same for either method, so the new potential energies are the same. 005 (part 2 of 2) 10.0 points What is the advantage of using the inclined plane? 1. less distance 2. less force correct 3. less total energy 4. more power Explanation: An inclined plane can exchange an in- creased distance for less force....
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This note was uploaded on 03/31/2009 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas at Austin.

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HW05-solutions - dang (ttd73) HW05 Tsoi (58020) 1 This...

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