Chapter 6, Problem 38P

Part (a)

To determine

Speed of artist when he lands

Answer to Problem 38P

Solution:

  $\text{v=6}\text{.26}\text{m/s}$

Explanation of Solution

Given Info:

Mass of artist is 62 kg.

Vertical speed of artist is 4.5 m/s.

The artist lands 2 m below the trampoline.

Spring constant of trampoline is $5.8\times {10}^4\text{N/m}$

Formula used: Kinetic energy of an object can be obtained by:

  $\text{KE=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}$

Potential Energy of spring can be obtained by:

  $\text{PE=}\frac{\text{1}}{\text{2}}{\text{Kx}}^{\text{2}}$

Where m is mass, v velocity, x is elongation, K is spring constant.

Gravitational potential energy of an object can be obtained by:

  $U=mgh$

Where, m is the mass, g is the gravitational acceleration and h is the height.

Calculation:

Applying conservation of energy, we equate the initial Potential energy to Kinetic energy when he lands on trampoline

  $\begin{array}{l}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{=mgh}\\ {\text{v}}^{\text{2}}\text{=2gh}\\ \text{v=}\sqrt{\text{2×9}\text{.8×2}}\\ \text{v=6}\text{.26m/s}\end{array}$

Conclusion:

The speed of artist is 6.26 m/s when he lands.

Part (b)

To determine

Depression of trampoline

Answer to Problem 38P

Solution:

  $\text{x=0}\text{.204 m}$

Explanation of Solution

Given:

Mass of artist is 62 kg.

Vertical speed of artist is 4.5 m/s.

The artist lands 2 m below the trampoline.

Spring constant of trampoline is $5.8\times {10}^4\text{N/m}$

Formula used: Kinetic energy of an object is given by

  $\text{KE=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}$

Potential Energy of spring is given by

  $\text{PE=}\frac{\text{1}}{\text{2}}{\text{Kx}}^{\text{2}}$

Where m is mass, v velocity, x is elongation, K is spring constant.

Calculation:

Applying conservation of energy, we equate the initial Potential energy to energy stored in trampoline at maximum depression

  $\begin{array}{l}\text{mgh=}\frac{\text{1}}{\text{2}}{\text{kx}}^{\text{2}}\\ \text{x=}\sqrt{\frac{\text{2mgh}}{\text{k}}}\end{array}$

Substituting values:

  $\begin{array}{l}\text{x=}\sqrt{\frac{\text{2×62×9}\text{.8×2}}{\text{5}{\text{.8×10}}^{\text{4}}}}\\ \text{x=0}\text{.204 m}\end{array}$

Conclusion:

From the above calculation the depression of trampoline is 0.204 m.