Physics Solution Manual for 1100 and 2101

There are two forces acting on it that are parallel

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Unformatted text preview: celeration and are, therefore, in equilibrium. According to Newton’s second law, the net force acting on each box must be zero. Thus, Newton’s second law applied to each box gives two equations in two unknowns, the magnitude of the tension in the rope between the boxes and the kinetic frictional force that acts on each box. Note that the frictional forces acting on the boxes are identical, because the boxes are identical. Solving these two equations shows that the tension is one-half of the applied force. 24. 31 kg·m/s2 25. 517 N 162 FORCES AND NEWTON'S LAWS OF MOTION CHAPTER 4 FORCES AND NEWTON'S LAWS OF MOTION PROBLEMS ______________________________________________________________________________ 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 = m PILOT m PLANE ΣF m PLANE ( ΣF )PILOT = mPILOT or Therefore, we find 3.7 × 104 N = 93 N 3.1 × 104 kg ______________________________________________________________________________ ( ΣF )PILOT = ( 78 kg ) 2. REASONING Suppose the bobsled is moving along the +x direction. There are two forces acting on it that are parallel to its motion; a force +Fx propelling it forward and a force of –450 N that is resisting its motion. The net force is the sum of these two forces. According to Newton’s second law, Equation 4.2a, the net force is equal to the mass of the bobsled times its acceleration. Since the mass and acceleration are known, we can use the second law to determine the magnitude of the propelling force. SOLUTION a. Newton’s second law states that + Fx − 450 N = ma x 14 244 4 3 ΣFx (4.2a) Solving this equation for Fx gives ( ) Fx = max + 450 N = ( 270 kg ) 2.4 m/s2 + 450 N = 1100 N b. The magnitude of the net force that acts on the bobsled is ( ) ΣFx = max = ( 270 kg ) 2.4 m/s2 = 650 N (4.2a) ____________________________________________________________________________________________ 3. REASONING In each case, we will apply Newton’s second law. Remember that it is the net force that appears in the second law. The net force is the vector sum of both forces. Chapter 4 Problems 163 SOLUTION a. We will use Newton’s second law, ΣFx = max, to find the force F2. Taking the positive x direction to be to the right, we have F1 + F2 = max 1 24 43 so F2 = max − F1 ΣFx F2 = (3.0 kg)(+5.0 m/s2) − (+9.0 N) = +6 N b. Applying Newton’s second law again gives F2 = max − F1 = (3.0 kg)(−5.0 m/s2) − (+9.0 N) = −24 N c. An application of Newton’s second law gives F2 = max − F1 = (3.0 kg)(0 m/s2) − (+9.0 N) = −9.0 N ______________________________________________________________________________ 4. 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 2.4), where the change in velocity is the final velocity v minus the initial velocity v0. SOLUTION The average net force exerted on the car and riders is v − v0 ) ( 45 m/s − 0 m/s = 5.5 × 103 kg = 3.5 × 104 N t − t0 7.0 s ______________________________________________________________________________ ∑ F = ma = m 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 2.9 from the equations of kinematics to determine the acceleration of the ball. SOLUTION According to Equation 2.9, the acceleration of the ball is given by a= 2 v 2 − v0 2x 164 FORCES AND NEWTON'S LAWS OF MOTION Thus, the magnitude of the net force on the ball is given by 2 v 2 − v0 (45 m/s)2 – (0 m/s)2 ∑ F = ma = m = (0.058 kg) = 130 N 2x 2(0.44 m) ______________________________________________________________________________ 6. REASONING AND SOLUTION The acceleration is obtained from x = v0t + 12 at 2 where v0 = 0 m/s. So a = 2x/t2 Newton’s second law gives 2 (18 m ) 2x = 2900 N ΣF = ma = m 2 = ( 72 kg ) ( 0.95 s )2 t ______________________________________________________________________________ 7. SSM REASONING AND SOLUTION The acceleration required is a= 2 v 2 − v0 2x − (15.0 m/s ) = = −2.25 m/s 2 2 ( 50.0 m ) 2 Newton's second law then gives the magnitude of the net force as F = ma = (1580 kg)(2.25 m/s2) = 3560 N ____________________________________________________________________________________________ 8. REASONING We do not have sufficient information to calculate the average net force applied to the fist by taking the vector sum of the individually applied forces. However, we have the mass m of the fist, as well as its initial velocity (v0 = 0 m/s, since the fist starts from rest), final velocity (v = 8.0 m/s), and the elapsed time (∆t = 0.15 s). Therefore we can use ∆v v − v0 Equation 2.4 a = = to determine the average acceleration a of the fist and then ∆t ∆t use Equation 4.1 (Newton’s second law,...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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