Chapter 6, Problem 39P

Part (a)

To determine

Upward speed of ball when released

Answer to Problem 39P

Solution:

  $\text{v= 7}\text{.47}\text{m/s}$

Explanation of Solution

Given:

Mass of ball is $0.380\text{ kg}.$

Compression is $0.160\text{ m}$ .

Spring constant is $875\text{ N/m}$ .

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

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

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 at original position to Kinetic energy and potential energy when it extends

  $\begin{array}{l}\frac{\text{1}}{\text{2}}{\text{Kx}}^{\text{2}}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{+mgh}\\ \text{v=}\sqrt{\frac{\frac{\text{1}}{\text{2}}{\text{Kx}}^{\text{2}}\text{-mgh}}{\frac{\text{1}}{\text{2}}\text{m}}}\end{array}$

Substituting values,

  $v=\sqrt{\frac{\frac{1}{2}875\times {(0.160)}^2-0.380\times 9.8\times 0.160}{\frac{1}{2}\times 0.380}}$

  $\text{v= 7}\text{.47}\text{m/s}$

Conclusion: From the above explanation, the speed of ball is $\text{v= 7}\text{.47}\text{m/s}$

Part (b)

To determine

Height above original position

Answer to Problem 39P

Solution:

  $\text{h=3 m}$

Explanation of Solution

Given:

Mass of ball is $0.380\text{ kg}.$

Compression is $0.160\text{ m}$ .

Spring constant is $875\text{ N/m}$ .

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 elastic potential energy at lowest position to gravitational potential energy at height h

  $\begin{array}{l}\text{mgh=}\frac{\text{1}}{\text{2}}{\text{Kx}}^{\text{2}}\\ \text{h=}\frac{{\text{kx}}^{\text{2}}}{\text{2mg}}\end{array}$

Substituting values,

  $\begin{array}{l}\text{h=}\frac{\text{875×(0}{\text{.160)}}^{\text{2}}}{\text{2×0}\text{.380×9}\text{.8}}\\ \text{h=3 m}\end{array}$

Conclusion:

From the above explanation, height above original position is 3 m.