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Unformatted text preview: homework 06 – – Due: Mar 1 2007, 4:00 am 1 Question 1, chap 6, sect 1. part 1 of 1 10 points A 930 kg car moves along a horizontal road at speed v = 22 m / s. The road is wet, so the static friction coefficient between the tires and the road is only μ s = 0 . 152 and the kinetic friction coefficient is even lower, μ k = 0 . 1064. 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: 162 . 46 m (tolerance ± 1 %). 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 = 162 . 46 m . Note: The answer does not depend on the car’s mass m . Question 2, chap 6, sect 1. part 1 of 2 6 points Consider a very small raindrop slowly falling down in quiet air. Because of the drop’s small size and low speed, the air re sistence to its fall is viscous rather than tur bulent. Consequently, the resistive force is depends linearly on the drop’s velocity, vector R = − bvectorv, (1) where b is a constant which depends on the size and shape of the drop and the air’s viscos ity but does not depend on the drop’s weight of what it’s made of. Suppose a raindrop of weight mg = . 043 μ g × 9 . 8 m / s 2 reaches terminal speed v t = 0 . 06 m / s high above the ground. What is the numeric value of the coefficient b in eq. (1) for the air resistence to this drop’s motion?...
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This note was uploaded on 01/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Friction, Work

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