Chapter 14, Problem 93A

(a)

To determine

The spring constant of spring.

Answer to Problem 93A

  $k=\text{2}{\text{.45×10}}^{\text{2}}\text{N/m}$

Explanation of Solution

Given:

Mass of spring: $\text{500}\text{.0g}$

Spring is compressed to: $\text{2}\text{.0}\text{cm}$

Formula used:

The formula to calculate the spring constant,$f=mg=kx$

Or,

  $k=\frac{mg}{x}$

Where,

  $k$ : Spring constant

  $F$ : Force

  $x$ : Length of String$g$ : Gravitational Force

  $m$ : Mass

Unit conversion:

  $\text{1cm = 0}\text{.01}\text{m}$

Hence,

  $\text{2}\text{.0 }\text{cm = 0}\text{.020}\text{m}$

  $\text{1}\text{g=0}\text{.001}\text{kg}$

Hence,

  $\text{500}\text{.0g=0}\text{.5}\text{kg}$

Calculation:

As the formula to calculate the spring constant is given by,

  $k=\frac{mg}{x}$

  $k=\frac{\left(0.5000\right)\times \left(9.8\right)}{0.020}$

  $k=\text{2}{\text{.45×10}}^{\text{2}}\text{N/m}$

Conclusion:

Hence, the spring constant is $k=\text{2}{\text{.45×10}}^{\text{2}}\text{N/m}$ .

(b)

To determine

If there is displacement of fish, the mass of fish.

Answer to Problem 93A

  $1.\text{125}\text{kg}$

Explanation of Solution

Given:

Mass of spring: $\text{500}\text{.0g}$

Fish displaces the spring: $\text{4}\text{.5}\text{cm}$

Formula used:

The formula to calculate the spring constant,$f=mg=kx$

Or also can be written as,

  $m=\frac{kx}{g}$

Where,

  $k$ : Spring constant

  $F$ : Force

  $x$ : Length of String$g$ : Gravitational Force

  $m$ : Mass

Unit conversion:

  $\text{1cm = 0}\text{.01 m}$

Hence,

  $4.5\text{cm}=4.5\times {10}^{-2}\text{m}$

Calculation:

To calculate the mass of spring if the fish displaces is given by formula,

  $m=\frac{kx}{g}$

Substituting the given values in the above formula,

  $m=\frac{\left(2.45\times {10}^2\text{N/m}\right)\times \left(4.5\times {10}^{-2}\right)}{9.8\text{N/kg}}$

  $m=1.\text{125}\text{kg}$

Conclusion:

Hence, the mass of the spring when the fish displaces is $1.\text{125}\text{kg}$ .