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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 6 / Problem 17E

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Answer: In Fig. E6.7 assume that there is no friction

Chapter 6, Problem 17E

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QUESTION:

In Fig. E6.7 assume that there is no friction force on the 20.0-N block that sits on the tabletop. The pulley is light and frictionless. (a) Calculate the tension T in the light string that connects the blocks. (b) For a displacement in which the 12.0-N block descends 1.20 m, calculate the total work done on (i) the 20.0-N block and (ii) the 12.0-N block. (c) For the displacement in part (b), calculate the total work done on the system of the two blocks. How does your answer compare to the work done on the 12.0-N block by gravity? (d) If the system is released from rest, what is the speed of the 12.0-N block when it has descended 1.20 m?

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QUESTION:

In Fig. E6.7 assume that there is no friction force on the 20.0-N block that sits on the tabletop. The pulley is light and frictionless. (a) Calculate the tension T in the light string that connects the blocks. (b) For a displacement in which the 12.0-N block descends 1.20 m, calculate the total work done on (i) the 20.0-N block and (ii) the 12.0-N block. (c) For the displacement in part (b), calculate the total work done on the system of the two blocks. How does your answer compare to the work done on the 12.0-N block by gravity? (d) If the system is released from rest, what is the speed of the 12.0-N block when it has descended 1.20 m?

ANSWER:

Solution 17E

Step 1:

(a).First we have to find masses to calculate tension.the tension formula is given by

$T=[({m}_{1}*{m}_{2})/({m}_{1}+{m}_{2})]*g$

${m}_{1}=20\ N/9.8\ m/{s}^{2}=2.04\ kg$

${m}_{2}=12\ N/9.8\ m/{s}^{2}=1.22\ kg$

$T=[(2.04\ kg*1.22\ kg)/(2.04\ kg+1.22\ kg)]*9.8\ m/{s}^{2}$

$T=[(0.76319)]*9.8\ m/{s}^{2}$

$T=7.4792$ ~ $7.5\ N.$

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