Q47E - University Physics with Modern Physics

Question

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of $\mathbf{40}.\mathbf{0}°$above the horizontal. The glider has mass $\mathbf{0}\mathbf{.}\mathbf{0900}\mathbf{ }\mathbf{kg}$ The spring has $\mathit{k}\mathbf{=}\mathbf{640}\mathbf{ }\mathbf{\text{N/m}}$ negligible mass. When the spring is released, the glider travels a maximum distance of $\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{ }\mathbf{m}$ along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. (a) What distance was the spring originally compressed? (b) When the glider has traveled along the air track $\mathbf{0}.\mathbf{80} \mathbf{m}$ from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

Step-by-Step Solution

Answer

(a) $0.056 \text{m}$

(b) $0.55 \text{J}$

Step 1: Identification of given data

Slope is $\theta =40.0°$

Mass of glider is $m=0.0900 \mathrm{kg}$$m=0.0900 \mathrm{kg}$

Spring constant is $k=640 \text{N/m}$

Maximum distance traveled by glider is $d=1.80 \mathrm{m}$

Step 2: Significance of law of conservation of energy

Energy cannot be created or destroyed; it can only be transformed from one form into another or transferred from one object to another, according to the physical rule of conservation of energy.

When an object is compressed or under strain, such as when a rubber band is stretched, elastic potential energy is stored (when you squeeze a spring). The kinetic energy, also known as the energy of motion, is created when the potential energy is "released."

$\text{Elastic} \text{Potentia} \text{Energy}\left(\frac{1}{2}k{x}^2\right)=\text{Potential} \text{Energy}\left(mgh\right)$                                                      ...(i)

Where, $k$ is spring constant, $x$ is the initial compression, $m$ is the mass of glider, $g$ is the acceleration due to gravity and $h$ is the height of the track

Step 3: (a) Determining the originally compressed spring

Height of track is given by

$\sin \theta =\frac{h}{d}$

Substitute the values in above equation

$\begin{array}{rcl}\sin 40.0°& =& \frac{h}{1.8 \text{m}}\\ h& =& 1.157 \text{m}\end{array}$

Substitute all the values in equation (i) to get the value of originally compressed spring

$\begin{array}{rcl}\frac{1}{2}k{x}^2& =& mgh\\ x& =& \sqrt{\frac{2mgh}{k}}\\ & =& \sqrt{\frac{2\times 0.09 \text{kg}\times 9.8 {\text{m/s}}^2\times 1.157 \text{m}}{640 \text{N/m}}}\\ & =& 0.056 \text{m}\end{array}$

Hence the value of originally compressed spring is $0.056 \text{m}$

Step 4: (b) Determining the kinetic energy of the glider

From part (a) we can see, that glider loses contact with the spring after$0.056 \text{m}$ so it not is in contact with the spring at point$0.056 \text{m}$

Given $d=0.80 \mathrm{m}$

kinetic energy of the glider at this point is given by 

$\text{KE}=\frac{1}{2}k{x}^2-mgd\sin \theta$

Substitute all the values in the above equation

$\begin{array}{rcl}\text{KE}& =& \frac{1}{2}\times 640 \text{N/m}\times {\left(0.056 \text{m}\right)}^2-\left(0.09 \text{kg}\times \text{9.8} {\text{m/s}}^2\times 0.8 \text{m}\right) \sin 40.0°\\ & =& 0.55 \text{J}\end{array}$

Hence the kinetic energy at this point is $0.55 \text{J}$