Chapter 11, Problem 66A

a.

To determine

Speed of the arrow when it leaves the bow.

Answer to Problem 66A

25.884 m/s

Explanation of Solution

Given:

Mass of arrow, $m=0.30\text{ kg}$

Average force, $F=201\text{ N}$

String is back by a distance, $s=0.5\text{ m}$

Formula used:

Work done is given by

  $W=F\cdot s$

Kinetic energy is given by,

  $KE=\frac{1}{2}m{v}^2$

According to work energy theorem.

  $F\cdot s=\frac{1}{2}m{v}^2$

Calculation:

Now, substitute the value of m, v, F and s and solve.

  $\begin{array}{l}201\times 0.5=\frac{1}{2}\times 0.3\times v^2\\ \frac{1}{2}\times 0.3\times v^2=100.5\\ v^2=670\\ v=25.884\text{ m/s}\end{array}$

Conclusion:

Therefore, speed of arrow when it leaves the bow is 25.884 m/s.

b

To determine

Arrow speed when it reaches the height of 10 m.

Answer to Problem 66A

14 m/s

Explanation of Solution

Given:

Mass of arrow, $m=0.30\text{ kg}$

Average force, $F=201\text{ N}$

String is back by a distance, $s=0.5\text{ m}$

Height, $h=10\text{ m}$

Formula used:

Kinetic energy is given by

  $KE=\frac{1}{2}m{v}^2$

Here, m is mass and v is velocity.

Potential energy is given by

  $PE=mgh$

Here, m is mass, g is acceleration due to gravity and h is height.

According to law of conservation of energy.

  $\frac{1}{2}m{v}^2=mgh$

Calculation:

Now, substituting the value of m, g and h and solve.

  $\begin{array}{l}\frac{1}{2}m{v}^2=mgh\\ \frac{1}{2}{v}^2=gh\\ \frac{1}{2}{v}^2=9.8\times 10\\ v^2=196\\ v=14\text{ m/s}\end{array}$

Conclusion:

Therefore, velocity of arrow will be $14\text{ m/s}$ .