Chapter 26, Problem 37P

(a)

To determine

The kinetic energy.

Answer to Problem 37P

  $\text{5}{\text{.5×10}}^{\text{19}}\text{J}$ .

Explanation of Solution

Given:

Mass, $\text{m =27000}\text{kg}$

Speed, $\text{v =}\text{0}\text{.21}\text{c}$

Formula used:

It is known that Kinetic energy,

  $KE=\left(m-m_0\right)c^2=m_0{c}^2\left(\left\{\frac{1}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}\right\}-1\right)$

Where,

  $m_0$ is rest mass

  $m$ is relativistic mass

  $v$ is velocity

c is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

Substitute the values

  $KE=\left(27,000kg\right){\left(3.00\times {10}^{8}m/s\right)}^2\left(\left\{\frac{1}{{\left[1-{\left(0.21\right)}^2\right]}^{\frac{1}{2}}}\right\}-1\right)$

  $KE=5.5\times {10}^{19}J$

(b)

To determine

The percentage error in kinetic energy.

Answer to Problem 37P

  $\text{-3}\text{.3}\text{\%}$ .

Explanation of Solution

Given:

Mass, $\text{m =27000}\text{kg}$

Speed, $\text{v =}\text{0}\text{.21}\text{c}$

Formula used:

It is known that Kinetic energy,

  $KE=\frac{1}{2}m{v}^2$

Where,

  $m$ is relativistic mass

  $v$ is velocity

Calculation:

Substitute the values

  $K{E}_1=\frac{1}{2}\left(27,000kg\right){\left(3.00\times {10}^{8}m/s\right)}^2=5.36\times {10}^{19}J$

The error is

  $E\%=\frac{\left(5.36-5.5\right)}{5.5}\times 100=-3.3\%$