Kaplunovsky - hw 06

# Kaplunovsky - hw 06 - homework 06 – – Due Mar 1 2007...

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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.

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Kaplunovsky - hw 06 - homework 06 – – Due Mar 1 2007...

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