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Unformatted text preview: boland (kbb544) – HW6 – mackie – (10611) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 5 . 65 kg block initially at rest is pulled to the right along a horizontal surface by a constant, horizontal force of 16 N. The coefficient of kinetic friction is 0 . 0914. The acceleration of gravity is 9 . 8 m / s 2 . Find the speed of the block after it has moved 3 . 35 m. Correct answer: 3 . 60168 m / s. Explanation: We have to use the equation Δ K = f s (1) , to calculate the change in the kinetic energy, Δ K . The net force exerted on the block is the sum of the applied 16 N force and the frictional force. Since the frictional force is in the direction opposite the displacement, it must be subtracted. The magnitude of the frictional force is f = μ N = μmg . Therefore the net force acting on the block is F net = F μmg = 16 N (0 . 0914) (5 . 65 kg) (9 . 8 m / s 2 ) = 10 . 9392 N . Multiplying this constant force by the dis placement, and using equation (1), we obtain Δ K = F net s = (10 . 9392 N) (3 . 35 m) = 36 . 6463 J = 1 2 mv 2 , since the initial velocity is zero. Therefore, v f = radicalbigg 2 Δ K m = radicalBigg 2 (36 . 6463 J) 5 . 65 kg = 3 . 60168 m / s . 002 10.0 points You leave your 100 W portable color TV on for 6 hours each day and you do not pay attention to the cost of electricity. If the dorm (or your parents) charged you for your electricity use and the cost was $0 . 2 / kW · h, what would be your monthly (30 day) bill? Correct answer: 3 . 6 dollars. Explanation: Let : P = 100 W and t = 6 h / day , The energy consumed in each day is W = P t = (100 W) (6 h / day) · kW 1000 W = 0 . 6 kW · h / day . In 30 days, you would use (30 day) (0 . 6 kW · h / day) = 18 kW · h , which would cost you (18 kW · h) ($0 . 2 / kW · h) = $3 . 6 . 003 10.0 points A car of weight 3190 N operating at a rate of 171 kW develops a maximum speed of 20 m / s on a level, horizontal road. Assuming that the resistive force (due to friction and air resistance) remains constant, what is the car’s maximum speed on an incline of 1 in 20; i.e. , if θ is the angle of the incline with the horizontal, sin θ = 1 / 20 ? Correct answer: 19 . 6337 m / s. Explanation: boland (kbb544) – HW6 – mackie – (10611) 2 If f is the resisting force on a horizontal road, then the power P is P = f v horizontal , so that f = P v h = (1 . 71 × 10 5 W) (20 m / s) = 8550 N . On the incline, the resisting force is F = f + mg sin θ = f + W 20 = P v h + W 20 . And, F v = P , so v = P F = P P v h + W 20 = (1 . 71 × 10 5 W) (1 . 71 × 10 5 W) (20 m / s) + (3190 N) 20 = 19 . 6337 m / s ....
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This note was uploaded on 02/20/2011 for the course PHYS 1062 taught by Professor Mackie during the Spring '11 term at Temple.
 Spring '11
 MACKIE
 Physics

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