This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 s = 0 . 136 and the kinetic friction coefficient is even lower, 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 = 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 = s N . Given s > k , it follows that static friction at its strongest is stronger than kinetic friction: The strongest possible friction force is F max f = s N which obtains only when the car does not skid. The normal force between the car and the road follows from Newtons 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 = 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 = s g. Now consider the kinematics of the cars deceleration to stop. Clearly, the shortest stopping distance obtains for the maximally negative a x , hence a x = a max = 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  = s g , hence the stopping distance X s = v 2 2 s g = 203 . 665 m . Note: The answer does not depend on the cars mass m . 002 (part 1 of 1) 8 points A block is at rest on an inclined plane. 7 4 k g s = . 3 6 c Find the critical angle, c , at which the block just begins to slide. Correct answer: 19 . 7989 . Explanation: Basic Concepts: Friction: f s s N , f k = k N Let : m = 74 kg , s = 0 . 36 , 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 c , the mag nitude of the static frictional force attains its maximum value 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 c =...
View Full
Document
 Spring '08
 Turner
 Physics

Click to edit the document details