solutions 10

solutions 10 - Patel Kinal – Homework 10 – Due Oct 9...

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Unformatted text preview: Patel, Kinal – Homework 10 – Due: Oct 9 2007, 7:00 pm – Inst: D Weathers 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A 3 . 27 g bullet moving at 842 m / s penetrates a tree to a depth of 3 . 87 cm. Use energy considerations to find the aver- age frictional force that stops the bullet. Correct answer: 29952 . 4 N. Explanation: We can use conservation of energy to relate the initial kinetic energy of the bullet to the work done by the frictional force. 1 2 mv 2 = f · s Solving for the frictional force, f , f = mv 2 2 s = (0 . 00327 kg)(842 m / s) 2 2(0 . 0387 m) = 29952 . 4 N . 002 (part 2 of 2) 10 points Assuming that the frictional force is constant, determine how much time elapsed between the moment the bullet entered the tree and the moment it stopped. Correct answer: 9 . 1924 × 10- 5 s. Explanation: If we assume the frictional force is constant, by Newton’s second law we can assume con- stant deceleration, which implies v avg = v initial 2 . The time of deceleration is then t = s v avg = . 0387 m 842 m / s 2 = 9 . 1924 × 10- 5 s . keywords: 003 (part 1 of 1) 10 points A car of weight 2160 N operating at power 161 kW develops a maximum speed of 32 . 8 m / s on a level, horizontal road. Assume: The resistive force remains con- stant under the conditions mentioned below. What is the maximum speed of the car up an incline of 9 . 18 ◦ relative to the horizontal? Correct answer: 30 . 6484 m / s. Explanation: Basic Concepts: P = dW dt = ~ F · ~v . Maximum Speed and Terminal Velocity Resistive forces. Let : W = 2160 N , p = 161 kW , v 1 = 32 . 8 m / s , and θ = 9 . 18 ◦ . Solution: On level ground, the only forces acting on the car in the horizontal direction are the applied force F 1 and the resistive force R . (Gravity and the normal force act perpen- dicular to this and add vectorially to zero.) Since the car has reached it maximum speed, it is no longer accelerating, so the net force is zero. That is F 1- R = 0 so R = F 1 . The applied force is related to the power through the relation P = ~ F 1 · ~v 1 . The ap- plied force, ~ F 1 , is in the same direction as the velocity, so ~ F 1 · ~v = F 1 v 1 . This means that F 1 = P v 1 = 161 kW 32 . 8 m / s µ 10 3 W kW ¶ = 4908 . 54 N . Patel, Kinal – Homework 10 – Due: Oct 9 2007, 7:00 pm – Inst: D Weathers 2 When the car is on the incline, we now have...
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solutions 10 - Patel Kinal – Homework 10 – Due Oct 9...

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