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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 6 / Problem 27E

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 little red wagon with mass 7.00 kg moves in a

Chapter 6, Problem 27E

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QUESTION:

A little red wagon with mass 7.00 kg moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 m/s and then is pushed 3.0 m in the direction of the initial velocity by a force with a magnitude of 10.0 N.

(a) Use the work–energy theorem to calculate the wagon’s final speed.

(b) Calculate the acceleration produced by the force. Use this acceleration in the kinematic relationships of Chapter 2 to calculate the wagon’s final speed. Compare this result to that calculated in part (a).

Questions & Answers

QUESTION:

A little red wagon with mass 7.00 kg moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 m/s and then is pushed 3.0 m in the direction of the initial velocity by a force with a magnitude of 10.0 N.

(a) Use the work–energy theorem to calculate the wagon’s final speed.

(b) Calculate the acceleration produced by the force. Use this acceleration in the kinematic relationships of Chapter 2 to calculate the wagon’s final speed. Compare this result to that calculated in part (a).

ANSWER:

Solution 27E

Step 1:

Data given

Mass $m=\ 7.00\ kg\$

Initial velocity ${v}_{i}=\ 4.00\ m/s$

Distance travelled $\Delta{}x=\ 3.00\ m\$

Force exerted $F=\ 10.0\ N$

We need to find the final speed of wagon wheel

Let us find the work done

The work done by the applied force over the 3.00 m distance

It is obtained using

$W=F\Delta{}x$

Substituting the values we get

$W=10.0\ N\ \times{}3.00\ m$

$W=\ 30.0\ \ J$

Thus we have work done = $30.0\ \ J$

As per the work energy relations

We can write

$\Sigma{}W=\ 1/2\ m{{v}_{i}}^{2}-1/2\ m{{v}_{f}}^{2}$

Substituting we get

$30\ J\ \ =\ 1/2\times{}\ 7.0\ kg\ \times{}{(4\ m/s)}^{2}-1/2\ \times{}7.0\ kg\ \times{}{{v}_{f}}^{2}$

Rearranging and solving we get

$v=\sqrt{\ \frac{2(30J)+7.00\ kg\ \times{}{(4\ m/s)}^{2}}{7.00\ kg\ }}$

$v=\sqrt{\ \frac{2((30J)+7.00\ kg\ \times{}{(4\ m/s)}^{2})}{7.00\ kg\ }}$

$v=\ 4.96\ m/s$

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