Chapter 10, Problem 95P

(a)

To determine

The spring constant of the spring.

Answer to Problem 95P

The spring constant of the spring is $98.0\text{N/m}$.

Explanation of Solution

The length of the spring is $20.0\text{cm}$, the mass of the brick hanged is $1.10\text{kg}$, the stretch of the spring is $31.0\text{cm}$.

Write the expression for total force in vertical y-direction on the spring.

$\begin{array}{c}{\displaystyle \sum F_y}=0\\ kd-mg=0\\ k=\frac{mg}{d}\end{array}$

Here, $k$ is the spring constant, $m$ is the mass hanged, $d$ is the stretched distance of the spring at equilibrium, $g$ is acceleration due to gravity.

Conclusion:

Substitute $1.10\text{kg}$ for $m$, $9.80{\text{m/s}}^{\text{2}}$ for $g$ and $31.0\text{cm}-20.0\text{cm}$ for $d$ to find the spring constant of the spring.

$\begin{array}{c}k=\frac{\left(1.10\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)}{31.0\text{cm}-20.0\text{cm}}\\ =\frac{\left(1.10\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)}{11.0\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}}\\ =98.0\text{N/m}\end{array}$

(b)

To determine

The maximum speed of the brick.

Answer to Problem 95P

The maximum speed of the brick is $0.472\text{m/s}$.

Explanation of Solution

The length of the spring is $20.0\text{cm}$, the mass of the brick hanged is $1.10\text{kg}$, the stretch of the spring is $31.0\text{cm}$, the pulled length of brick from equilibrium point is $5.00\text{cm}$.

Write the expression for conservation of energy to find the maximum speed of the brick.

$\begin{array}{c}{K}_{\max }=U_{\max }\\ \frac{1}{2}m{v}_{\max }^2=\frac{1}{2}k{d}^2\\ v_{\max }=\sqrt{\frac{k{d}^2}{m}}\end{array}$

Here, $K_{\max }$ is the maximum kinetic energy, $U_{\max }$ is the maximum potential energy, $v_{\max }$ is the maximum speed of the brick.

Conclusion:

Substitute $1.10\text{kg}$ for $m$, $98.0\text{N/m}$ for $k$ and $5.00\text{cm}$ for $d$ to find the maximum speed of the brick.

$\begin{array}{c}{v}_{\max }=\sqrt{\frac{\left(98.0\text{N/m}\right){\left(5.00\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}^2}{1.10\text{kg}}}\\ =0.472\text{m/s}\end{array}$

(c)

To determine

The speed of the brick when the displacement is $2.50\text{cm}$.

Answer to Problem 95P

The speed of the brick when the displacement is $2.50\text{cm}$ is $0.409\text{m/s}$.

Explanation of Solution

The length of the spring is $20.0\text{cm}$, the mass of the brick hanged is $1.10\text{kg}$, the stretch of the spring is $31.0\text{cm}$.

Write the expression for conservation of energy to find the speed of the brick.

$\begin{array}{c}E=K+U\theta \\ \frac{1}{2}k{d}^2=\frac{1}{2}m{v}^2+\frac{1}{2}k{y}^2\\ v=\sqrt{\frac{k}{m}\left(d^2-y^2\right)}\end{array}$

Here, $y$ is the displacement of the brick.

Conclusion:

Substitute $1.10\text{kg}$ for $m$, $98.0\text{N/m}$ for $k$, $5.00\text{cm}$ for $d$ and $2.50\text{cm}$ for $y$ to find the speed of the brick.

$\begin{array}{c}v=\sqrt{\frac{98.0\text{N/m}}{1.10\text{kg}}\left({\left(5.00\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}^2-{\left(2.50\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}^2\right)}\\ =0.409\text{m/s}\end{array}$

(d)

To determine

The time taken for the brick to oscillate five times.

Answer to Problem 95P

The time taken for the brick to oscillate five times is $3.33\text{s}$.

Explanation of Solution

The length of the spring is $20.0\text{cm}$, the mass of the brick hanged is $1.10\text{kg}$, the stretch of the spring is $31.0\text{cm}$.

Write the expression for time taken for the brick to oscillate five times.

$5T=5\left(2\pi \sqrt{\frac{m}{k}}\right)$

Conclusion:

Substitute $1.10\text{kg}$ for $m$, $98.0\text{N/m}$ for $k$ to find the time taken for the brick to oscillate five times.

$\begin{array}{c}5T=5\left(2\pi \sqrt{\frac{1.10\text{kg}}{98.0\text{N/m}}}\right)\\ =10\pi \sqrt{\frac{1.10\text{kg}}{98.0\text{N/m}}}\\ =3.33\text{s}\end{array}$