Physx141MPch4sol

Physx141MPch4sol - 4.2. Visualize: Assess: Note that the...

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4.2. Visualize: Assess: Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse walls. 4.4. Model: Assume friction is negligible compared to other forces. Visualize: 4.5. Visualize: Please refer to Figure Ex4.5. Solve: Mass is defined to be 1 slope of the acceleration-versus-force graph m = A larger slope implies a smaller mass. We know m 2 = 0.20 kg, and we can find the other masses relative to m 2 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 m 3 > m 2 and m 1 < m 2 . This is consistent with our numerical answers. 4.6. 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 2.4 m/s 2 . (b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2 m ). Using , Fm a = 2 F = (2 m ) a a = F / m = 0.6 m/s 2
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4.8.
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This note was uploaded on 05/05/2008 for the course PHYS 141 taught by Professor Staff during the Spring '06 term at Cal Poly.

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Physx141MPch4sol - 4.2. Visualize: Assess: Note that the...

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