2 Chapter and 3 Questions
1. Chapter 2, Question 9 (6th edition): You throw a ball straight up from the edge of a cliff, and it lends on the ground below the cliff. If you had, instead, thrown the ball down from the cliff edge with the same speed, would the ball's speed just before landing be larger than, smaller than, or the same as previously? (Hint: Consider Eq. 2-16.) The ball's speed just before landing would be the same as previously. When the ball is thrown vertically upward with the initial speed vo it comes back to the same height having the same speed vo and moving downward. So, when one throws the ball directly downward with the speed vo (instead upward first), the final speed at the bottom of the cliff is the same as when it was thrown with the same initial speed upward. 2. Chapter 3, Question 2 (6th edition): Can the magnitude of the difference between two vectors ever be greater than (a) the magnitude of one of the vectors, (b) the magnitudes of both vectors, and (c) the magnitude of their sum? The magnitude of the difference (or the sum) between two vectors A and B has always the value between the difference of their magnitudes and the sum of their magnitudes: || A | - | B || | A B | | A | + | B | Therefore, (a) the magnitude of the difference between two vectors can be larger than the magnitude of one of the vectors; (b) can be larger than magnotudes of both vectors; and (c) can be larger than the magnitude of their sum.
A @ A A@ A @ A @ B-A B+A A @ AB @ @ A @ R A U ? -
A
Z A A A B A A U A -
(a)
C
?
(b)
C
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Z B Z Z A
~ Z >
(c)
A C = B - A, | C |<| B | and | C |>| A |
C = A - B, | C |>| A | and | C |>| B |
A
-A
| B - A |>| B + A |
Chapter 5 and 6 Questions
1. Chapter 5, Question 3: In Fig. 5-23 (shown below), forces F1 and F2 are applied to a lunch box as it slides at constant velocity over a frictionless floor. We are to decrease angle without changing the magnitude of F1 . To keep the lunch box sliding at constant velocity, should we increase, decrease, or remain the magnitude of F2 ? First, the lunch box slides with constant velocity, which means that the net force acting on the box is zero, and therefore, the x-component of the force F1 is equal to and opposite to the force F2 . By decreasing the angle we are increasing the magnitude of the x-component of the force F1 . To keep the lunch box sliding at constant velocity we should increase the magnitude of F2 so that the net force remains zero.
' F2 B F1
2. Chapter 5, Question 8 (6th edition): A vertical force F is applied to a block of mass m that lies on a floor. What happens to the magnitude of the normal force FN on the block from the floor as magnitude |F | is increased from zero if force F is (a) downward and (b) upward? When the magnitude of the vertical force F increases, the magnitude of the normal force FN will (a) increase if the vertical force F is downward, and (b) bf decrease if the vertical force F is upward. 3. Chapter 6, Question 2 (6th edition): In figure (a) below, a thermos is sent sliding leftward across a long plastic tray. What are the directions of the kinetic frictional force on (a) the thermos and (b) the tray from each other? (c) Does the kinetic frictional force increase or decrease speed the of the thermos relative to the floor? In figure (b) below, the tray is now sent sliding leftward beneath the thermos. What now are the directions of the kinetic frictional forces on (d) the thermos and (e) the tray from each other? (f) Does the kinetic frictional force increase or decrease the speed of the thermos relative to the floor? (g) Do kinetic frictional forces always slow objects? In figure (a) the thermos is sent sliding leftward across the tray. The direction of the kinetic frictional force on (a) the thermos is toward right or against the motion and (b) on the tray is toward left, representing the reaction force in an action-reaction pair. (c) Given the initial push, the thermos would move with constant velocity v relative to the floor if there would not be any friction. With the friction between the thermos and tray, after the initial push the frictional force is going to decrease the thermos' speed relative to the floor. In figure (b) the tray is sent sliding leftward beneath the thermos. Now, the direction of the kinetic frictional force on (d) the thermos is toward left representing the reaction force in an action-reaction pair, and (e) the kinetic friction on the tray is toward right or against motion. (f) Given the initial push, the thermos would not move at all relative to the floor if there would not be any friction. With the friction between thermos and tray present, the initial push is going to produce a motion of the thermos in the direction of motion of the tray, therefore the kinetic friction would increase the speed of the thermos relative to the floor. (g) Without frictional force the thermos would not move in the example shown in figure (b), so it represents an example of kinetic frictional force causing the actual motion. Therefore, kinetic frictional forces do not always slow objects.
' v
' v
(a)
(b)
4. Chapter 6, Question 5: In the figure b...