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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 6 / Problem 14E

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: A 1.50-kg book is sliding along a rough horizontal

Chapter 6, Problem 14E

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QUESTION:

Problem 14E

A 1.50-kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s, and at point B it has slowed to 1.25 m/s. (a) How much work was done on the book between A and B? (b) If - 0.750 J of work is done on the book from B to C, how fast is it moving at point C? (c) How fast would it be moving at C if + 0.750 J of work was done on it from B to C?

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QUESTION:

Problem 14E

A 1.50-kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s, and at point B it has slowed to 1.25 m/s. (a) How much work was done on the book between A and B? (b) If - 0.750 J of work is done on the book from B to C, how fast is it moving at point C? (c) How fast would it be moving at C if + 0.750 J of work was done on it from B to C?

ANSWER:

Solution 14E

Step 1:

We need to find work done between point A and B

It is given by

Using

$K=\ 1/2\ m{v}^{2}$

Since we have two points we shall find the difference in kinetic energy between A and B

We have

${v}_{A}=3.21\ m/s\$

${v}_{B}=1.25\ m/s\$

$m=\ 1.50\ kg\$

Kinetic energy at A

${K}_{A}=1/2\ m{{v}_{A}}^{2}\$

${K}_{A}=(1/2)\times{}1.50\ kg\times{}(3.21\ m/s{)}^{2}\ \$

${K}_{A}=\ 7.72\ J$

Similarly

Kinetic energy at B

${K}_{B}=1/2\ m{{v}_{B}}^{2}\$

${K}_{B}=(1/2)\times{}1.50\ kg\times{}(1.25\ m/s{)}^{2}\ \$

${K}_{B}=\ 1.17\ J$

We get work done as

$W=\ {K}_{B}-{K}_{A}$

$W=\ 1.17\ J-7.72\ J$

$W=\ -6.55\ J\$

Hence we have work done as $\ -6.55\ J$

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