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Unformatted text preview: Nguyen, Phan – Midterm 2 – Due: Mar 10 2004, 10:00 pm – Inst: V Kaplunovsky 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 7 points A 1200 kg car moves along a horizontal road at speed v = 23 . 3 m / s. The road is wet, so the static friction coefficient between the tires and the road is only μ s = 0 . 136 and the kinetic friction coefficient is even lower, μ k = 0 . 0952. The acceleration of gravity is 9 . 8 m / s 2 . What is the shortest possible stopping dis tance for the car under such conditions? Use g = 9 . 8 m / s 2 and neglect the reaction time of the driver. Correct answer: 203 . 665 m. Explanation: The force stopping the car is the friction force F f between the tires and the road. If the car skids , the friction force is governed by the kinetic friction law, F f = μ k N where N is the normal force between the car and the road. If the car does does not skid , the friction is static and the friction force can be anything up to a maximal static friction F max f = μ s N . Given μ s > μ k , it follows that static friction at its strongest is stronger than kinetic friction: The strongest possible friction force is F max f = μ s N which obtains only when the car does not skid. The normal force between the car and the road follows from Newton’s Laws for the ver tical direction of motion: Since the car moves in the horizontal direction only , a y = 0 and hence F net y = N  mg = 0 . Consequently, N = mg , F max f = μ s mg, and since there are no horizontal forces other than friction, the acceleration is limited to  a x  ≤ a max = 1 m F max f = μ s g. Now consider the kinematics of the car’s deceleration to stop. Clearly, the shortest stopping distance obtains for the maximally negative a x , hence a x = a max = μ s g = const . At constant deceleration, the stopping time of the car is t s = v  a x  , which gives us the stopping distance X s = v t s  a x  2 t 2 s = v 2 2  a x  . For the problem at hand,  a x  = μ s g , hence the stopping distance X s = v 2 2 μ s g = 203 . 665 m . Note: The answer does not depend on the car’s mass m . 002 (part 1 of 1) 8 points A block is at rest on an inclined plane. 7 4 k g μ s = . 3 6 θ c Find the critical angle, θ c , at which the block just begins to slide. Correct answer: 19 . 7989 ◦ . Explanation: Basic Concepts: Friction: f s ≤ μ s N , f k = μ k N Let : m = 74 kg , μ s = 0 . 36 , θ c = 19 . 7989 ◦ , and v f = final speed . F f N mg 19 . 7989 ◦ Nguyen, Phan – Midterm 2 – Due: Mar 10 2004, 10:00 pm – Inst: V Kaplunovsky 2 Solution: At the critical angle θ c , the mag nitude of the static frictional force attains its maximum value μ s N . Since the block is at rest, the component of the gravitational force parallel to the incline is equal in magnitude to the frictional force mg sin θ c =...
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This note was uploaded on 03/23/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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