Q45E - University Physics with Modern Physics

Question

At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large compressed spring. The spring, with force constant $\mathit{k}\mathbf{=}\mathbf{40}\mathbf{ }\mathbf{\text{N/cm}}$ and negligible mass, rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass $\mathbf{70}\mathbf{ }\mathbf{\text{kg}}$  are pushed against the other end, compressing the spring $\mathbf{0}\mathbf{.}\mathbf{375}\mathbf{ }\mathbf{\text{m}}$ . The sled is then released with zero initial velocity. What is the sled’s speed when the spring (a) returns to its uncompressed length and (b) is still compressed $\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{ }\mathbf{\text{m}}$?

Step-by-Step Solution

Answer

1. The speed of the sled when spring returns to uncompressed length is $2.8 \text{m/s}$.
2. The velocity of the sled when the spring is still compressed is $2.4 \text{m/s}$
   

Step 1: Identification of given data

The given data can be listed below,

 

The force constant of the spring is, $\text{k}=40 \text{N/cm}$

The total mass of rider and sled is, $\text{m}=70 \text{kg}$

The compressed length of the spring is, $\text{x}=0.375 \text{m}$

Step 2: Concept/Significance of elastic potential energy,

Kinetic energy is the power an item or body possesses as a result of motion. Every particle engaged in motion of one type or another possesses it.

Step 3: (a) Determination of the sled’s speed when the spring returns to its uncompressed length.

The maximum potential energy of the spring when it is compressed is given by,

${\mathrm{U}}_{\mathrm{el}}=\frac{1}{2}{\mathrm{kx}}^2$

Here, k is the force constant of the spring and x is the compressed length of the spring.

 

Substitute all the values in the above,

$\begin{array}{rcl}{\mathrm{U}}_{\mathrm{el}}& =& \frac{1}{2}\left(4000 \text{N/m}\right){\left(0.375 \text{m}\right)}^2\\ & =& 20 \text{N/m}\left(0.1406 {\text{m}}^2\right)\\ & =& 281 \text{N}·\text{m}\end{array}$

According to the work-energy theorem, the speed of the sled when spring returns to uncompressed length is given by,

$$\begin{gathered} {\mathrm{U}}_{\mathrm{el}}=\mathrm{\Delta K} \\ {\mathrm{U}}_{\mathrm{el}}=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{v}=\sqrt{\frac{2{\mathrm{U}}_{\mathrm{el}}}{\mathrm{m}}} \end{gathered}$$

Here,  ${\mathrm{U}}_{\mathrm{el}}$ is the potential energy of the spring, and m is the mass of the system.

 

Substitute all the values in the above,

$\begin{array}{rcl}\mathrm{v}& =& \sqrt{\frac{2\times 281 \text{N}·\text{m}}{70 \text{kg}}}\\ & =& \sqrt{\frac{281 \text{N}·\text{m}}{35 \text{kg}}\left(\frac{1 \text{kg}·{\text{m/s}}^2}{1 \text{N}}\right)}\\ & =& 2.8 \text{m/s}\end{array}$

Thus, the speed of the sled when spring returns to uncompressed length is $2.8 \text{m/s}$.

Step 4: (b) Determination of the sled’s speed when the spring is still compressed 0 . 200   m .

The potential energy of the spring when it is compressed further for $0.200 \text{m}$ is given by,

$\mathrm{U}=\frac{1}{2}\mathrm{k}{\left(\mathrm{\delta x}\right)}^2$

Here, k is the force constant of spring and $\mathrm{\delta x}$  is the final compressed length of the spring.

 

Substitute all the values in the above,

$\begin{array}{rcl}U& =& \frac{1}{2}\left(4000 \text{N/m}\right){\left(0.375 \text{m}-0.200 \text{m}\right)}^2\\ & =& 2000 \text{N/m}{\left(0.175 \text{m}\right)}^2\\ & =& 201 \text{J}\end{array}$

By the work-energy theorem, the velocity of the sled is given by,

$$\begin{gathered} \mathrm{U}=\mathrm{\Delta K} \\ \mathrm{U}=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{v}=\sqrt{\frac{2\mathrm{U}}{\mathrm{m}}} \end{gathered}$$

Substitute all the values in the above,

$\begin{array}{rcl}\mathrm{v}& =& \sqrt{\frac{2\times 201 \text{J}}{70 \text{kg}}}\\ & =& \sqrt{5.742 {\left(\text{m/s}\right)}^2}\\ & =& 2.4 \text{m/s}\end{array}$

Thus, the velocity of the sled when the spring is still compressed is $2.4 \text{m/s}$.