2. REASONING AND SOLUTION When the birdfeeder is hanging freely and no one is pulling on the dangling (lower) cord, there is a tension in the cord

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1 . REASONING AND SOLUTION When the birdfeeder is hanging freely and no one is pulling on the dangling (lower) cord, there is a tension in the cord between the birdfeeder and the tree limb (the upper cord), because the upper cord supports the weight of the birdfeeder. When the lower cord is pulled down with a slow continuous pull, the tension in both cords increases slowly. Since the upper cord has a larger tension to begin with, it always has the greater tension as the lower cord is pulled. Thus, the upper cord snaps first. On the other hand, when the child gives the lower cord a sudden, downward pull, the tension in the lower cord increases suddenly. However, the tension in the upper cord does not increase as suddenly. The reason is that the birdfeeder has a large mass, so it accelerates very slowly. Thus, the upper cord is stretched slowly and, consequently, the tension in the upper cord rises slowly. Since the tension rises much faster in the lower cord, it is the first to snap.

2 3. REASONING AND SOLUTION If the net external force acting on an object is zero, it is possible for the object to be traveling with a nonzero velocity. According to Newton s second law, ΣF = ma, if the net external force ΣF is zero, the acceleration a is also zero. If the acceleration is zero, the velocity must be constant, both in magnitude and in direction. Thus, an object can move with a constant nonzero velocity when the net external force is zero.

3 4. REASONING AND SOLUTION According to Newton's second law, a net force is required to give an object a non-zero acceleration. a. If an object is moving with a constant acceleration of 9.80 m/s, we can conclude that there is a net force on the object. b. If an object moves with a constant velocity of 9.80 m/s, its acceleration is zero; therefore, we can conclude that the net force acting on the object is zero.

4 1. REASONING AND SOLUTION According to Newton s second law, the acceleration is a = ΣF/m. Since the pilot and the plane have the same acceleration, we can write ΣF ΣF or ( F) m ΣF = Σ = PILOT m m m PILOT PILOT PLANEPLANE Therefore, we find N Σ = 78 kg 93 N PILOT 4 = kg ( F ) ( )

5 . REASONING Newton s second law of motion gives the relationship between the net force F and the acceleration a that it causes for an object of mass m. The net force is the vector sum of all the external forces that act on the object. Here the external forces are the drive force, the force due to the wind, and the resistive force of the water. SOLUTION We choose the direction of the drive force (due west)as the positive direction. Σ F = ma for the acceleration gives Solving Newton s second law ( ) Σ F N 800 N 100 N a = = = m/s m 6800 kg The positive sign for the acceleration indicates that its direction is due west.

6 3. REASONING According to Newton s second law, Equation 4.1, the average net force ΣF is equal to the product of the object s mass m and the average acceleration a. The average acceleration is equal to the change in velocity divided by the elapsed time (Equation.4), where the change in velocity is the final velocity v minus the initial velocity v 0. SOLUTION The average net force exerted on the car and riders is v v F ma m t t ( ) 3 45 m/s 0 m/s kg N 0 = = = = s

7 4. REASONING AND SOLUTION Using Equation.4 and assuming that t 0 = 0 s, we have for the required time that t = v v 0 a Since ΣF = ma, it follows that ( ) t = v v 0 ΣF / m = m v v 0 ΣF [ ( )] ( 5.0 kg) ( m/s) 0 m/s = N = s

8 5. SSM REASONING The net force acting on the ball can be calculated using Newton's second law. Before we can use Newton's second law, however, we must use Equation.9 from the equations of kinematics to determine the acceleration of the ball. SOLUTION According to Equation.9, the acceleration of the ball is given by a = v v 0 x Thus, the magnitude of the net force on the ball is given by v v0 (45 m/s) (0 m/s) F = ma = m = (0.058 kg) = 130 N x (0.44m)

9 6. REASONING AND SOLUTION The acceleration required is a= v -v 0 x = - (15.0 m/s) (50.0 m) = -.5 m/s Newton's second law then gives the magnitude of the net force as F = ma = (1580 kg)(.5 m/s ) = 3560 N

10 7. SSM REASONING According to Newton's second law of motion, the net force applied to the fist is equal to the mass of the fist multiplied by its acceleration. The data in the problem gives the final velocity of the fist and the time it takes to acquire that velocity. The average acceleration can be obtained directly from these data using the definition of average acceleration given in Equation.4. SOLUTION The magnitude of the average net force applied to the fist is, therefore, v F ma m ( 0.70 kg) 8.0m/s 0m/s = = = = 37 N t 0.15 s

11 8. REASONING AND SOLUTION From Equation.9, v = v 0 + ax Since the arrow starts from rest, v 0 = 0 m/s. In both cases, x is the same so v 1 v = a 1x a x = a 1 a or v 1 v = a 1 a Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and or v = v 1 F F 1 = v 1 F 1 F 1 v 1 v = F 1 F = ( 5.0 m/s) = 35.4 m/s

12 9. SSM WWW REASONING Let due east be chosen as the positive direction. Then, when both forces point due east, Newton's second law gives F + F = ma A B 1 (1) ΣF where a 1 = 0.50 m/s. When F A points due east and F B points due west, Newton's second law gives where F F = ma A B () ΣF a = 0.40 m/s. These two equations can be used to find the magnitude of each force. SOLUTION a. Adding Equations 1 and gives F A ( + a ) ( 8.0kg)( 0.50m/s m/s ) m a1 = = = 3.6 N b. Subtracting Equation from Equation 1 gives ( a ) ( 8.0kg)( 0.50m/s 0.40m/s ) m a1 FB = = = 0.40 N

5. REASONING AND SOLUTION An object will not necessarily accelerate when two or more forces are applied to the object simultaneously.

5. REASONING AND SOLUTION An object will not necessarily accelerate when two or more forces are applied to the object simultaneously. The applied forces may cancel so the net force is zero; in such a case,