Chapter 7, Problem 36P

(a)

To determine

To Find: The force constant of the spring.

Answer to Problem 36P

  $14.2N/m$

Explanation of Solution

Given:

Mass is $0.250kg$ .

The mass is dropped from rest at point A, with the spring initially unstretched. As the mass falls, the spring stretches and reaches point B.

Formula Used:

The potential energy of the object is given as

  $PE=mgh$

Where,

  $m=$ Mass

  $g=$ Acceleration due to gravity

  $h=$ Height

The potential energy of spring is given as

  $P{E}_{Spring}=\frac{1}{2}K{x}^2$

Where,

  $K=$ Spring constant

  $x=$ Distance moved

Calculation:

Using the conservation of energy, energy at point A is equal to energy at point B

  $\begin{array}{l}{\left(PE\right)}_A={\left(KE\right)}_A+{\left(P{E}_{Spring}\right)}_A\\ mgh=\frac{1}{2}m{v}^2+\left(\frac{1}{2}K{x}^2\right)\\ mgh=\frac{1}{2}m{v}^2-\frac{1}{2}K{x}^2\end{array}$

Spring constant is derived as

  $\begin{array}{l}2mgh=m{v}^2+K{x}^2\\ K{x}^2=2mgh-m{v}^2\\ K=\frac{2mgh-m{v}^2}{x^2}\end{array}$

Here,

  $\begin{array}{l}m=0.250kg\\ g=9.8m/s^2\\ v=5.00m/s\\ x=2.00m-1.50m=0.50m\\ h=2.00m\end{array}$

Substituting the values in above formula

  $\begin{array}{l}K=\frac{2\left(0.25\right)\left(9.8\right)\left(2\right)-\left(0.25\right){\left(5\right)}^2}{{\left(0.5\right)}^2}\\ K=\frac{9.8-6.25}{0.25}\\ K=\frac{3.55}{0.25}\\ K=14.2N/m\end{array}$

Conclusion:

Thus, force constant of the spring is $14.2N/m$

(b)

To determine

To Find: The magnitude of the acceleration of the mass at point B

Answer to Problem 36P

  $18.6m/s^2$

Explanation of Solution

Given:

Mass is $0.250kg$ .

The mass is dropped from rest at point A, with the spring initially unstretched. As the mass falls, the spring stretches and reaches point B.

Formula Used:

According to Newton’s second law of motion, force is equal to mass times acceleration

  $F=ma$

Where,

  $m=$ Mass

  $a=$ Acceleration

Calculation:

Using Newton’s second law

  $F_{Net}=ma$

Also, $F_{Net}=kx-mg$

Thus, equating the above equations

  $kx-mg=ma$

On solving

  $a=\frac{kx-mg}{m}$

Here,

  $\begin{array}{l}k=14.2N/m\\ m=0.250kg\\ x=2.00m-1.50m=0.50m\\ g=9.8m/s^2\end{array}$

Substituting the value in equation

  $\begin{array}{l}a=\frac{\left(14.2\right)\left(0.5\right)-\left(0.250\right)\left(9.8\right)}{\left(0.250\right)}\\ a=\frac{7.1-2.45}{0.25}\\ a=\frac{4.65}{0.25}\\ a=18.6m/s^2\end{array}$

Conclusion:

Thus, magnitude of the acceleration of the mass at point B is $18.6m/s^2$ .