Chapter 5

# Chapter 5 - 5.1. Visualize: Assess: walls. Note that the...

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5.1. Visualize: Assess: Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse walls.

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5.2. Visualize:
5.3. Visualize:

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5.4. Model: Assume friction is negligible compared to other forces. Visualize:
5.5. Visualize: Assess: The bow and archer are no longer touching the arrow, so do not apply any forces after the arrow is released.

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5.6. Model: An object’s acceleration is linearly proportional to the net force. Solve: (a) One rubber band produces a force F , two rubber bands produce a force 2 F , and so on. Because Fa and two rubber bands (force 2 F ) produce an acceleration of 1.2 m/s 2 , four rubber bands will produce an acceleration of 2 2.4 m/s . (b) Now, we have two rubber bands (force 2 F ) pulling two glued objects (mass 2 m ). Using , Fm a = 2 2( 2 ) /0 . 6 m / s a a F m =⇒ = =
5.7. Solve: Let the object have mass m and each rubber band exert a force F . For two rubber bands to accelerate the object with acceleration a , we must have 2 . F a m = We will need N rubber bands to give acceleration 3 a to a mass 1 . 2 m Find N : 1 2 22 33 3 . NF F NF aN mm m ⎛⎞ = ⇒= = ⎜⎟ ⎝⎠ Three rubber bands are required.

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5.8. Visualize: Please refer to Figure EX5.8. Solve: Mass is defined to be 1 slope of the acceleration-versus-force graph m = A larger slope implies a smaller mass. We know 2 0.20 kg, m = and we can find the other masses relative to 2 m by comparing their slopes. Thus 1 2 12 1/slope 1 slope 2 1 2 0.40 1/slope 2 slope 1 5 2 5 0.40 0.40 0.20 kg 0.08 kg m m mm == = = = ⇒= = × = Similarly, 3 2 32 1/slope 3 slope 2 1 5 2.50 1/slope 2 slope 3 2 5 2 2.50 2.50 0.20 kg 0.50 kg m m = = = = Assess: From the initial analysis of the slopes we had expected > and . < This is consistent with our numerical answers.
5.9. Visualize: Please refer to Figure EX5.9. Solve: Mass is defined to be 1 slope of the acceleration-versus-force graph m = Thus () 1 1 1 1 11 2 1 5 3 slope of line 1 9 3 5 5 25 slope of line 2 3 3 5 a m N m a N −− ⎛⎞ ⎜⎟ ⎝⎠ == = = The ratio of masses is 1 2 9 25 m m = Assess: More rubber bands produce a smaller acceleration on object 2, so it should be more massive.

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5.10 Solve: Use proportional reasoning. Given that distance traveled is proportional to the square of the time, 2 , dt so 2 d t should be constant. We have () 22 2.0 furlongs 2.0 s 4.0 s x = Thus the distance traveled by the object in 4.0 s is x = 8.0 furlongs. Assess: A longer time should result in a longer distance traveled.
5.11 Solve: Use proportional reasoning. Let period T = of the pendulum, length L = of pendulum. We are told , TL so T L should be constant. We have 3.0 s 2.0 m 3.0 m x = Solving, the period of the 3.0 m long pendulum is 3.7 s. x = Assess: Increasing the length increases the period, as expected.

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5.12. Force is not necessary for motion. Constant velocity motion occurs in the absence of forces, that is, when the net force on an object is zero. Thus, it is incorrect to say that “force causes motion.” Instead, force causes acceleration . That is, force causes a change in the motion of an object, and acceleration is the kinematic quantity that measures a change of motion. Newton’s second law quantifies this idea by stating that the net force
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## This note was uploaded on 11/26/2010 for the course MTH 232 taught by Professor Smith during the Spring '10 term at UConn.

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Chapter 5 - 5.1. Visualize: Assess: walls. Note that the...

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