Chapter 7, Problem 1Q

(a)

To determine

To find: the kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=4\widehat{i}+3\widehat{j}$ velocity.

Answer to Problem 1Q

The kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=4\widehat{i}+3\widehat{j}$ velocity is $5m$ .

Explanation of Solution

Given:

  $\stackrel{\rightharpoonup }{v}=4\widehat{i}+3\widehat{j}$

Calculation:

The kinetic energy of a particle can be determined using the equation of kinetic energy.

The kinetic energy:

  $\begin{array}{l}KE=\frac{1}{2}m{v}^2\\ KE=\frac{1}{2}m{\left(\sqrt{4^2+3^2}\right)}^2\\ KE=\frac{1}{2}m\left(25\right)\\ KE=5m\end{array}$

Thus, the kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=4\widehat{i}+3\widehat{j}$ velocity is $5m$ .

(b)

To determine

To find: the kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=-4\widehat{i}+3\widehat{j}$ velocity.

Answer to Problem 1Q

The kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=-4\widehat{i}+3\widehat{j}$ velocity is $5m$ .

Explanation of Solution

Given:

  $\stackrel{\rightharpoonup }{v}=-4\widehat{i}+3\widehat{j}$

Calculation:

The kinetic energy of a particle can be determined using the equation of kinetic energy.

The kinetic energy:

  $\begin{array}{l}KE=\frac{1}{2}m{v}^2\\ KE=\frac{1}{2}m{\left(\sqrt{-4^2+3^2}\right)}^2\\ KE=\frac{1}{2}m\left(25\right)\\ KE=5m\end{array}$

Thus, the kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=-4\widehat{i}+3\widehat{j}$ velocity is $5m$ .

(c)

To determine

To find: the kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=-3\widehat{i}+4\widehat{j}$ velocity.

Answer to Problem 1Q

The kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=-3\widehat{i}+4\widehat{j}$ velocity is $5m$ .

Explanation of Solution

Given:

  $\stackrel{\rightharpoonup }{v}=-3\widehat{i}+4\widehat{j}$

Calculation:

The kinetic energy of a particle can be determined using the equation of kinetic energy.

The kinetic energy:

  $\begin{array}{l}KE=\frac{1}{2}m{v}^2\\ KE=\frac{1}{2}m{\left(\sqrt{-3^2+4^2}\right)}^2\\ KE=\frac{1}{2}m\left(25\right)\\ KE=5m\end{array}$

Thus, the kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=-3\widehat{i}+4\widehat{j}$ velocity is $5m$ .

(d)

To determine

To find: the kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=3\widehat{i}-4\widehat{j}$ velocity.

Answer to Problem 1Q

The kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=3\widehat{i}-4\widehat{j}$ velocity is $5m$ .

Explanation of Solution

Given:

  $\stackrel{\rightharpoonup }{v}=3\widehat{i}-4\widehat{j}$

Calculation:

The kinetic energy of a particle can be determined using the equation of kinetic energy.

The kinetic energy:

  $\begin{array}{l}KE=\frac{1}{2}m{v}^2\\ KE=\frac{1}{2}m{\left(\sqrt{3^2+{\left(-4\right)}^2}\right)}^2\\ KE=\frac{1}{2}m\left(25\right)\\ KE=5m\end{array}$

Thus, the kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=3\widehat{i}-4\widehat{j}$ velocity is $5m$ .

(e)

To determine

To find: the kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=5\widehat{i}$ velocity.

Answer to Problem 1Q

The kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=5\widehat{i}$ velocity is $5m$ .

Explanation of Solution

Given:

  $\stackrel{\rightharpoonup }{v}=5\widehat{i}$

Calculation:

The kinetic energy of a particle can be determined using the equation of kinetic energy.

The kinetic energy:

  $\begin{array}{l}KE=\frac{1}{2}m{v}^2\\ KE=\frac{1}{2}m{\left(\sqrt{5^2}\right)}^2\\ KE=\frac{1}{2}m\left(25\right)\\ KE=5m\end{array}$

Thus, the kinetic energy of a particle with $\stackrel{\rightharpoonup }{v}=5\widehat{i}$ velocity is $5m$ .

(f)

To determine

To find: the kinetic energy of a particle with $v=5$ at $30°$ .

Answer to Problem 1Q

The kinetic energy of a particle with $v=5$ at $30°$ is $5m$ .

Explanation of Solution

Given:

  $v=5$ at $30°$

Calculation:

The kinetic energy of a particle can be determined using the equation of kinetic energy.

The kinetic energy:

  $\begin{array}{l}KE=\frac{1}{2}m{v}^2\\ KE=\frac{1}{2}m{\left(\sqrt{5^2\left({\cos }^{2}30+{\sin }^{2}30\right)}\right)}^2\\ KE=\frac{1}{2}m\left(25\left(1\right)\right)\\ KE=5m\end{array}$

Thus, the kinetic energy of a particle with $v=5$ at angle $30°$ is $5m$ .

Since, all particle’s kinetic energy is same.

Hence, the rating of the kinetic energy is: a=b=c=d=e=f.