Microquiz_4

# Microquiz_4 - ConcepTest 5.2 Antilock Brakes Antilock...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ConcepTest 5.2 Antilock Brakes Antilock brakes keep the car wheels from locking and skidding during a sudden stop. Why does this help slow the car down? 1) k > s so sliding friction is better 2) k > s so static friction is better 3) s > k so sliding friction is better 4) s > k so static friction is better ConcepTest 5.2 Antilock Brakes Antilock brakes keep the car wheels from locking and skidding during a sudden stop. Why does this help slow the car down? 1) k > s so sliding friction is better 2) k > s so static friction is better 3) s > k so sliding friction is better 4) s > k so static friction is better Static friction is greater than sliding friction, so friction by keeping the wheels from skidding, the static friction force will help slow the car down more efficiently than the sliding friction that occurs during a skid. ConcepTest 5.3 Going Sledding Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this? 1) pushing her from behind 2) pulling her from the front 3) both are equivalent 1 2 ConcepTest 5.3 Going Sledding Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this? In Case 1, the force F is pushing down (in addition to mg), so the normal force is larger. In Case 2, the force F larger is pulling up, against gravity, so the up normal force is lessened. Recall that lessened the frictional force is proportional to the normal force. 1) pushing her from behind 2) pulling her from the front 3) both are equivalent 1 2 ConcepTest 5.5b Sliding Down II A mass m is placed on an inclined plane ( > 0) and slides down the plane with constant speed. If a similar block (same ) of mass 2m were placed on the same incline, it would: 1) not move at all 2) slide a bit, slow down, then stop 3) accelerate down the incline 4) slide down at constant speed m ConcepTest 5.5b Sliding Down II A mass m is placed on an inclined plane ( > 0) and slides down the plane with constant speed. If a similar block (same ) of mass 2m were placed on the same incline, it would: 1) not move at all 2) slide a bit, slow down, then stop 3) accelerate down the incline 4) slide down at constant speed The component of gravity acting down the plane is double for 2m. However, the normal force (and hence the friction force) is also double (the same factor!). This means the two forces still cancel to give a net force of zero. f Wy N W Wx ConcepTest 5.7b Around the Curve II During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening? (1) centrifugal force is pushing you into the door (2) the door is exerting a leftward force on you (3) both of the above (4) neither of the above ConcepTest 5.7b Around the Curve II During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening? (1) centrifugal force is pushing you into the door (2) the door is exerting a leftward force on you (3) both of the above (4) neither of the above The passenger has the tendency to continue moving in a straight line. There is a centripetal force, provided by the door, that forces the passenger into a circular path. ConcepTest 5.9 Ball and String Two equal-mass rocks tied to strings are whirled in horizontal circles. The radius of circle 2 is twice that of circle 1. If the period of motion is the same for both rocks, what is the tension in cord 2 compared to cord 1? 1) T2 = 1/4 T1 2) T2 = 1/2 T1 3) T2 = T1 4) T2 = 2 T1 5) T2 = 4 T1 ConcepTest 5.9 Ball and String Two equal-mass rocks tied to strings are whirled in horizontal circles. The radius of circle 2 is twice that of circle 1. If the period of motion is the same for both rocks, what is the tension in cord 2 compared to cord 1? 1) T2 = 1/4 T1 2) T2 = 1/2 T1 3) T2 = T1 4) T2 = 2 T1 5) T2 = 4 T1 The centripetal force in this case is given by the tension, so T = mv2/r. For the same period, we find that v2 = 2v1 (and this term is squared). However, for the denominator, we see that r2 = 2r1 which gives us the relation T2 = 2T1. ConcepTest 5.10 Barrel of Fun A rider in a "barrel of fun" finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? 1 2 3 4 5 ConcepTest 5.10 Barrel of Fun A rider in a "barrel of fun" finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? 1 2 3 4 5 The normal force of the wall on the rider provides the centripetal force needed to keep her going around in a circle. The downward force of gravity is balanced by the upward frictional force on her, so she does not slip vertically. Follow-up: What happens if the rotation of the ride slows down? ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online