Physics Solution Manual for 1100 and 2101

The problem text gives the electrons initial velocity

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Unformatted text preview: ΣF = ma ) to find the average net force ΣF applied to the fist. SOLUTION Inserting the relation a = average net force applied to the fist: ∆v v − v0 = into Newton’s second law yields the ∆t ∆t Chapter 4 Problems 165 v − v0 8.0 m/s − 0 m/s ΣF = ma = m = ( 0.70 kg ) = 37 N 0.15 s ∆t 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 + FB = ma1 1A24 43 ΣF (1) where a1 = 0.50 m/s 2 . When FA points due east and FB points due west, Newton's second law gives F – FB = ma2 1A24 43 ΣF (2) where a2 = 0.40 m/s 2 . These two equations can be used to find the magnitude of each force. SOLUTION a. Adding Equations 1 and 2 gives FA = m ( a1 + a2 ) 2 = (8.0 kg ) ( 0.50 m / s2 + 0.40 m / s 2 ) 2 = 3.6 N b. Subtracting Equation 2 from Equation 1 gives FB = m ( a1 − a2 ) 2 = (8.0 kg ) ( 0.50 m / s2 − 0.40 m / s 2 ) 2 = 0.40 N ____________________________________________________________________________________________ 10. REASONING From Newton’s second law, we know that the net force ΣF acting on the electron is directly proportional to its acceleration, so in part a we will first find the electron’s acceleration. The problem text gives the electron’s initial velocity (v0 = +5.40×105 m/s) and final velocity (v = +2.10×106 m/s), as well as its displacement (x = +0.038 m) during the interval of acceleration. The elapsed time is not known, so we ( ) 2 will use Equation 2.9 v 2 = v0 + 2ax to calculate the electron’s acceleration. Then we will find the net force acting on the electron from Equation 4.1 ( ΣF = ma ) and the electron’s mass. Because F1 points in the +x direction and F2 points in the −x direction, the net force 166 FORCES AND NEWTON'S LAWS OF MOTION acting on the electron is ΣF = F1 − F2 . In part b of the problem, we will rearrange this expression to obtain the magnitude of the second electric force. SOLUTION a. Solving Equation 2.9 for the electron’s acceleration, we find that a= 2 v 2 − v0 2x ( +2.10 ×106 m/s ) − ( +5.40 ×105 m/s ) = 2 2 2 ( +0.038 m ) = +5.42 × 1013 m/s 2 Newton’s 2nd law of motion then gives the net force causing the acceleration of the electron: ∑ F = ma = ( 9.11×10−31 kg ) ( +5.42 ×1013 m/s2 ) = +4.94 ×10−17 N b. The net force acting on the electron is ΣF = F1 − F2 , so the magnitude of the second electric force is F2 = F1 − ΣF , where ΣF is the net force found in part a: F2 = F1 − ΣF = 7.50 × 10−17 N − 4.94 ×10−17 N = 2.56 × 10−17 N 11. REASONING Newton’s second law gives the acceleration as a = (ΣF)/m. Since we seek only the horizontal acceleration, it is the x component of this equation that we will use; ax = (ΣFx)/m. For completeness, however, the free-body diagram will include the vertical forces also (the normal force FN and the weight W). The free-body diagram is SOLUTION shown at the right, where +y F1 = 59.0 N F2 = 33.0 N θ = 70.0° When F1 is replaced by its x and y components, we obtain the free body diagram in the following drawing. FN F2 +x θ W F1 Chapter 4 Problems Choosing right to be the positive direction, we have ΣF F cos θ − F2 ax = x = 1 m m ax ( 59.0 N ) cos 70.0° − ( 33.0 N ) = = 7.00 kg −1.83 m/s 167 +y FN 2 F2 The minus sign indicates that the horizontal acceleration points to the left . F1cos θ +x F1sin θ W 12. REASONING The net force ΣF has a horizontal component ΣFx and a vertical component ΣFy. Since these components are perpendicular, the Pythagorean theorem applies (Equation 1.7), and the magnitude of the net force is ΣF = ( ΣFx ) 2 () + ΣFy 2 . Newton’s second law allows us to express the components of the net force acting on the ball in terms of its mass and the horizontal and vertical components of its acceleration: ΣFx = ma x , ΣFy = ma y (Equations 4.2a and 4.2b). SOLUTION Combining the Pythagorean theorem with Newton’s second law, we obtain the magnitude of the net force acting on the ball: ΣF = ( ΣFx ) 2 = ( 0.430 kg ) 13. () + ΣFy 2 = ( max ) 2 ( + ma y ) 2 (810 m/s2 ) + (1100 m/s2 ) 2 2 2 = m ax + a 2 y = 590 N SSM REASONING To determine the acceleration we will use Newton’s second law ΣF = ma. Two forces act on the rocket, the thrust T and the rocket’s weight W, which is mg = (4.50 × 105 kg)(9.80 m/s2) = 4.41 × 106 N. Both of these forces must be considered when determining the net force ΣF. The direction of the acceleration is the same as the direction of the net force. SOLUTION In constructing the free-body diagram for the rocket we choose upward and to the right as the positive directions. The free-body diagram is as follows: 168 FORCES AND NEWTON'S LAWS OF MOTION The x component of the net force is ΣFx = T cos 55.0° = 7.50 ×106 N cos 55.0° = 4.30 ×106 N ( T +y Ty ) 55.0º The y component of the net force is W ( = T sin 55.0º +x Tx = T cos 55.0º ) ΣFy = T sin 55.0° − W = 7.50 × 106 N sin 55.0° − 4.41×106 N = 1.73 × 106 N The magnitudes of th...
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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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