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Unformatted text preview: dang (ttd73) – HW03 – Tsoi – (58020) 1 This printout should have 30 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1 . 5 s. A passenger in the elevator is holding a 9 . 5 kg bundle at the end of a vertical cord. The acceleration of gravity is 9 . 8 m / s 2 . What is the tension in the cord as the ele vator accelerates? Correct answer: 101 . 544 N. Explanation: T mg a elevator g Let h be the distance traveled and a the acceleration of the elevator. With the initial velocity being zero, we simplify the following expression and solve for acceleration of the elevator: h = v t + 1 2 at 2 = 1 2 at 2 = ⇒ a = 2 h t 2 . The equation describing the forces acting on the bundle is F net = ma = T − mg T = m ( g + a ) = m parenleftbigg g + 2 h t 2 parenrightbigg = (9 . 5 kg) bracketleftbigg 9 . 8 m / s 2 + 2 (1 m) (1 . 5 s) 2 bracketrightbigg = 101 . 544 N . 002 (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 D 2. W = + μ k ( N + mg cos θ ) D 3. W = − μ k ( N + mg cos θ ) D 4. W = + μ k ( N − mg cos θ ) D 5. W = − μ k N D correct 6. W = − μ k ( N − mg cos θ ) D 7. W = 0 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. 003 (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 = ( mg cos θ + F sin θ −N ) D 3. W = −N D 4. W = 0 correct 5. W = N D dang (ttd73) – HW03 – Tsoi – (58020) 2 6. W = ( N − mg cos θ − F sin θ ) D 7. W = ( N + mg cos θ + F sin θ ) 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. 004 (part 3 of 3) 10.0 points What is the final speed of the block? 1. v = radicalbigg 2 m ( F cos θ − μ k N ) D 2. v = radicalbigg 2 m ( F cos θ − mg sin θ − μ k N ) D correct 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 − μ k N ) D 5. v = radicalbigg 2 m ( F sin θ − μ k N ) D 6. v = radicalbigg 2 m ( F sin θ + μ k N ) D 7. v = radicalbigg 2 m ( F cos θ − mg sin θ ) D 8. v = radicalbigg 2 m ( F cos θ + mg sin θ ) D Explanation: The work done by gravity is W grav = mg D cos(90 ◦ + θ ) = − mg D sin θ ....
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 Spring '08
 Kaplunovsky
 Force, Friction, kg, Fnet

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