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Unformatted text preview: cerutti (cpc566) – HW05 – TSOI – (58160) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 0.0 points WITHDRAWN 002 (part 2 of 2) 0.0 points WITHDRAWN 003 10.0 points A cheerleader lifts his 21 . 8 kg partner straight up off the ground a distance of 0 . 302 m before releasing her.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 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 + mg cos θ ) D 3. W = + μ k ( N  mg cos θ ) D 4. W = 0 5. W = μ k N D correct 6. W = + μ k N D 7. W = μ k ( N  mg 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. 005 (part 2 of 3) 10.0 points What is the work done by the normal force N ? 1. W = N D sin θ 2. W = N D 3. W = 0 correct 4. W = ( N  mg cos θ F sin θ ) D 5. W = ( mg cos θ + F sin θ N ) D 6. W =N D 7. W = N D cos 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. 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 θ mg sin θ + μ k N ) D 2. v = radicalbigg 2 m ( F sin θ μ k N ) D 3. v = radicalbigg 2 m ( F cos θ + mg sin θ μ k N ) D 4. v = radicalbigg 2 m ( F cos θ mg sin θ ) 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 μ k N ) D 8. v = radicalbigg 2 m ( F sin θ + μ k N ) D Explanation: The work done by gravity is W grav = mg D cos(90 ◦ + θ ) = mg D sin θ . 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 . Thus v f = radicalbigg 2 m ( F cos θ mg sin θ μ 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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 Spring '08
 Turner
 Physics, Force, Correct Answer, partner, Mg 3, H1

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