hw5 - cerutti (cpc566) HW05 TSOI (58160) 1 This print-out...

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Unformatted text preview: cerutti (cpc566) HW05 TSOI (58160) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 0.0 points WITHDRAWN 002 (part 2 of 2) 0.0 points WITHDRAWN 003 10.0 points A cheerleader lifts his 21 . 8 kg partner straight up off the ground a distance of 0 . 302 m before releasing her. The acceleration of gravity is 9 . 8 m / s 2 . If he does this 29 times, how much work has he done? Correct answer: 1871 . 06 J. Explanation: The work done in lifting the cheerleader once is W 1 = mg h = (21 . 8 kg)(9 . 8 m / s 2 )(0 . 302 m) = 64 . 5193 J . The work required to lift her n = 29 times is W = nW 1 = (29)(64 . 5193 J) = 1871 . 06 J . 004 (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 + mg cos ) D 2. W = + k ( N + mg cos ) D 3. W = + k ( N - mg cos ) D 4. W = 0 5. W =- k N D correct 6. W = + k N D 7. W =- k ( N - mg cos ) D 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. 005 (part 2 of 3) 10.0 points What is the work done by the normal force N ? 1. W = N D sin 2. W = N D 3. W = 0 correct 4. W = ( N - mg cos - F sin ) D 5. W = ( mg cos + F sin - N ) D 6. W =-N D 7. W = N D cos 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. 006 (part 3 of 3) 10.0 points What is the final speed of the block? cerutti (cpc566) HW05 TSOI (58160) 2 1. v = radicalbigg 2 m ( F cos - mg sin + k N ) D 2. v = radicalbigg 2 m ( F sin - k N ) D 3. v = radicalbigg 2 m ( F cos + mg sin - k N ) D 4. v = radicalbigg 2 m ( F cos - mg sin ) D 5. v = radicalbigg 2 m ( F cos - mg sin - k N ) D correct 6. v = radicalbigg 2 m ( F cos + mg sin ) D 7. v = radicalbigg 2 m ( F cos - k N ) D 8. v = radicalbigg 2 m ( F 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 . 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 . 007 10.0 points Five ramps lead from the ground to the second floor of a workshop, as sketched below. All five ramps have the same height; ramps B, C, D and E have the same length; ramp A is longer than the other four. You need to push a heavy cart up to the second floor and you may choose any one of the five ramps....
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hw5 - cerutti (cpc566) HW05 TSOI (58160) 1 This print-out...

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