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aguilar (fa6754) – hk7 – opyrchal – (11106) This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A(n) 561 kg elevator starts from rest. It moves upward for 4.73 s with a constant acceleration until it reaches its cruising speed of 1.97 m/s. The acceleration of gravity is 9.8 m/s2 . Find the average power delivered by the elevator motor during the period of this acceleration. 1 002 10.0 points A block of mass 0.21 kg is placed on a vertical spring of constant 3227 N/m and pushed downward, compressing the spring 0.15 m. After the block is released it leaves the spring and continues to travel upward. The acceleration of gravity is 9.8 m/s2 . What height above the point of release will the block reach if air resistance is negligible? Correct answer: 17.6403 m. Explanation: Correct answer: 5.64548 kW. Explanation: Given : Let : m = 561 kg , t = 4.73 s , and v = 1.97 m/s , The height y is y = vave t 1 = vt 2 1 = (1.97 m/s) (4.73 s) 2 = 4.65905 m . Since the elevator starts from rest, the power P supplied by the motor, totally transferred into Kinetic Energy K and Potential Energy U , where E = K + U , is U +K E = t t 1 m g y + 2 m v2 = t 1 m g 2 v t + 12 m v 2 = t mv = (g t + v) 2t (561 kg) (1.97 m/s) = 2 (4.73 s) × [(9.8 m/s2 ) (4.73 s) + 1.97 m/s] = 5645.48 W = 5.64548 kW . P = m = 0.21 kg , x = 0.15 m , k = 3227 N/m , g = 9.8 m/s2 . and Choose Ug = 0 at the level of the release point. Ki = Kf = 0, so from conservation of energy, (Ug + Us )i = (Ug + Us )f 0 + Us,i = Ug,f + 0 1 k x2 = m g h 2 k x2 h= 2mg (3227 N/m)(0.15 m)2 = 2(0.21 kg)(9.8 m/s2 ) = 17.6403 m . 003 (part 1 of 2) 10.0 points A pendulum consists of a sphere of mass 1.7 kg attached to a light cord of length 11.1 m as in the figure below. The sphere is released from rest when the cord makes a 41.9◦ angle with the vertical, and the pivot at P is frictionless. The acceleration of gravity is 9.8 m/s2 . aguilar (fa6754) – hk7 – opyrchal – (11106) 2 For the velocity at the lowest point, we obtain 11 .1 m p vb = 2 g L (1 − cos θi ) h = 2 (9.8 m/s2 ) (11.1 m) i1/2 × (1 − cos 41.9◦ ) 1.7 kg 41.9◦ = 7.45839 m/s , 2 9.8 m/s Find the speed of the sphere when it is at the lowest point. Correct answer: 7.45839 m/s. Explanation: Correct answer: 25.1795 N. Let : m = 1.7 kg , ℓ = 11.1 m , θi = 41.9◦ , and g = 9.8 m/s2 . The only force that does work on the sphere is the force of gravity, since the force of tension is always perpendicular to each element of the displacement and hence does no work. Since the force of gravity is a conservative force, the total mechanical energy is constant. Therefore, as the pendulum swings, there is a continuous transfer between potential and kinetic energy. At the instant the pendulum is released, the energy is entirely potential energy. At the lowest point, the pendulum has kinetic energy but has lost some potential energy. The pendulum swings to the opposite side and stops at the same height as it started. The pendulum has regained its potential energy and its kinetic energy is again zero. If we measure the y coordinates from the center of rotation, then ya = −L cos θ 004 (part 2 of 2) 10.0 points What is the tension of the cord at the lowest point? and yb = −L . and Ub = −m g L . Explanation: Since the force of tension does no work, it cannot be determined using the energy method. To find Tb , we can apply Newton’s second law to the radial direction. First, recall that the centripetal acceleration of a particle moving in a circle of radius L is equal v2 directed toward the center of rotation. to L Therefore, X m vb2 Fr = Tb − m g = m ar = . L Substituting the expression for vb obtained in the first part of the problem, into the Newton’s second law, we find for the tension at point the lowest point Tb = m g + 2 m g (1 − cos θi ) = m g (3 − 2 cos θi ) = (1.7 kg) (9.8 m/s2 ) (3 − 2 cos 41.9◦ ) = 25.1795 N . Therefore, i Ua = −m g L cos θi Applying the principle of constancy of mechanical energy gives K a + Ua = K b + Ub 1 0 − m g L cos θi = m vb2 − m g L . 2 005 10.0 points A spring with a spring-constant 2.6 N/cm is compressed 24 cm and released. The 9 kg mass skids down the frictional incline of height 43 cm and inclined at a 24◦ angle. The acceleration of gravity is 9.8 m/s2 . aguilar (fa6754) – hk7 – opyrchal – (11106) The path is frictionless except for a distance of 0.5 m along the incline which has a coefficient of friction of 0.4 . k = 2.6 N/cm 9 kg 43 cm 24 cm 0. 24◦ µ = 0. vf Figure: Not drawn to scale. What is the final velocity vf of the mass? Correct answer: 2.55165 m/s. Kf = Us + Ul − Wf r = (7.488 J) + (37.926 J) − (16.1149 J) = 29.2991 J . Kf = 2 Let : g = 9.8 m/s = , k = 2.6 N/cm = 260 N/m , x = 24 cm = 0.24 m , µ = 0.4 , ℓ = 0.5 m , h = 0.43 m , m = 9 kg , and θ = 24◦ , 1 m v2 . 2 (3) Multiplying by 2 and dividing by m gives us 2 Kf = v 2 , so m r 2 Kf v= s m = Consider the kinetic energy of the mass. The mass receives its initial kinetic energy from the potential energy of the spring (1) It gains kinetic energy because of the potential energy lost in moving down the incline Kgained = Ulost = mgh (2) 2 = (9 kg) (9.8 m/s ) (0.43 m) = 37.926 J . and loses kinetic energy by doing work on the frictional surface Klost = Wf r (3) However, the final kinetic energy is Explanation: Ki = Uspring 1 = k x2 2 1 = (260 N/m) (0.24 m)2 2 = 7.488 J . = µ m g ℓ cos θ = (0.4) (9 kg) (9.8 m/s2 ) × (0.5 m) cos(24◦ ) = 16.1149 J . Since energy is concerved, the final kinetic energy is 5m 4 3 2 (29.2991 J) (9 kg) = 2.55165 m/s . Alternate Explanation: The potential energy at the top of the hill will be converted into kinetic energy at the bottom of the hill minus energy lost due to the nonconservative friction force. The potential energy at the top of the hill consists of the gravitational potential energy plus the potential energy contained in the compressed spring. Combining Eqs. 1, 2, 3, and 4, we have s 2 1 2 v= m g h + k x − µ m g ℓ cos θ m 2 006 (part 1 of 2) 10.0 points The spring constant of a toy dart gun is 2934 N/m. To cock the gun the spring is compressed 1.1 cm. The 10 g dart, fired straight upward, reaches a maximum height of 23 m. aguilar (fa6754) – hk7 – opyrchal – (11106) 4 For the return trip, k0 = 0 , vf = 0 , so Wext = ∆K + ∆U + Wa = 0 23 m 1.1 cm Determine the magnitude of the energy dissipated by air friction during the dart’s ascent. The acceleration of gravity is 9.81 m/s2 . 1 m vf 2 − m g h + Wa = 0 . 2 Since 2 (m g h − Wa ) m 2 (0.01 kg) (9.81 m/s2 ) (23 m) = 0.01 kg 2 (2.07879 J) − 0.01 kg = 35.5014 m2 /s2 , vf2 = Correct answer: 2.07879 J. Explanation: Let : k = 2934 N/m , x = 1.1 cm = 0.011 m , m = 10 g = 0.01 kg , h = 23 m , and g = 9.81 m/s2 . Applying conservation of energy during the dart’s ascent (k0 = kf = 0 , v0 = 0) , Wext = ∆K + ∆U + Wa = 0 1 − k x2 + m g h − Wa = 0 2 1 k x2 2 = (0.01 kg) (9.81 m/s2 ) (23 m) 1 − (2934 N/m) (0.011 m)2 2 = 2.07879 J . vf = q 35.5014 m2 /s2 = 5.95831 m/s . 008 (part 1 of 3) 10.0 points A 18 kg block is dragged over a rough, horizontal surface by a constant force of 131 N acting at an angle of 32 ◦ above the horizontal. The block is displaced 15.7 m, and the coefficient of kinetic friction is 0.125. The acceleration of gravity is 9.8 m/s2 . 13 Wa = m g h − 18 kg µ = 0.125 Find the work done by the 131 N force. and the energy dissipated is 2.07879 J . Correct answer: 1744.18 J. 007 (part 2 of 2) 10.0 points What is the maximum speed the projectile could have when it returns to its starting point? Assume the difference in the frictional energy loss due to the diminished velocity on the way down compared to its velocity on the way up is negligible. Explanation: Correct answer: 5.95831 m/s. Explanation: 1N ◦ 32 Let : m = 18 kg , F = 131 N , sx = 15.7 m , µ = 0.125 , and g = 9.8 m/s2 . Consider the force diagram aguilar (fa6754) – hk7 – opyrchal – (11106) 5 If the block was originally at rest, determine its final speed. n F θ fk Correct answer: 13.0564 m/s. Explanation: The net work done on the block is equal to the change in kinetic energy, so mg ~ · ~s, where ~s is the distance Work is W = F traveled. In this problem ~s = 5ı̂ is only in the x direction. 1 m v 2 − 0 = WF + Wµ 2 2 [WF + Wµ ] v h m i u u 2 (1744.18 J) + (−209.949 J) t = (18 kg) v= ⇒ WF = Fx sx = F sx cos θ = (131 N) (15.7 m) cos 32◦ = 1744.18 J . 009 (part 2 of 3) 10.0 points Find the work done by the force of friction. Correct answer: −209.949 J. Explanation: To find the frictional force, Ff riction = µ N , we need to find N from vertical force balance. Note that N is in the same direction as the y component of F and opposite the force of gravity. Thus F sin θ + N = m g so that N = m g − F sin θ . Thus the friction force is ~ f riction = −µ N ı̂ = −µ (m g − F sin θ)ı̂ . F The work done by friction is then ~ f riction · ~s = −|fµ | |s| Wµ = F = −µ (m g − fµ sin θ) sx = −(0.125) [(18 kg) (9.8 m/s2 ) −(131 N) sin 32◦ ] (15.7 m) = −209.949 J . 010 (part 3 of 3) 10.0 points r = 13.0564 m/s .