Q6-40E - University Physics with Modern Physics

Question

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do$\mathbf{80}\mathbf{.0}\mathbf{\text{\hspace{0.17em}J}}$ of work when you compress the springs$\mathbf{0}\mathbf{.200}\mathbf{\text{\hspace{0.17em}m}}$ from their uncompressed length. 

(a) What magnitude of force must you apply to hold the platform in this position? 

(b) How much additional work must you do to move the platform$\mathbf{0}\mathbf{.200}\mathbf{\text{\hspace{0.17em}m}}$ farther, and what maximum force must you apply?

 

Step-by-Step Solution

Answer

a) The magnitude of force applied to hold the platform in position is.$800\text{\hspace{0.17em}N}$

b) The additional work done to move the platform and maximum force applied on the platform are $240\text{\hspace{0.17em}J}$and $1600\text{\hspace{0.17em}N}$respectively.

 

Step 1: Identification of given data

The given data can be listed below,

• The work done on the spring to compress it is,$W=80\text{\hspace{0.17em}J}$
• The compressed length of the spring is,$x_2=0.200\text{\hspace{0.17em}m}$

 

Step 2: Concept/Significance of spring constant

A measurement of the spring force's elasticity is the spring constant. The spring constant is the ratio of the force of the spring to the extra length that results from hanging a mass vertically in the spring.

Step 3: (a) Determination of the magnitude of force applied to hold the platform in the position

The free body diagram of the system is shown below as,

Fromthe work done equation, the force constant of the spring is given by,

 

$\begin{array}{c}W=\frac{1}{2}k{x}_2^2-\frac{1}{2}k{x}_1^2\\ k=\frac{2W}{x_2^2-x_1^2}\end{array}$

 

Here, W is the force constant of the spring,$x_2$ is the compressed length of the spring,$x_1$ is the initially compressed length.

 

On substituting all the values, the force constant of spring is found as,

 $\begin{array}{c}k=\frac{2\times 80\text{\hspace{0.17em}J}}{{\left(0.200\text{\hspace{0.17em}m}\right)}^2-{\left(0\right)}^2}\\ =4000\text{\hspace{0.17em}N/m}\end{array}$

 

The force on the spring is given by,

 

$F=k{x}_2$

 

Here, k is the force constant of the spring, $x_2$is the compressed length of the spring.

 

Substitute all the values in the above,

 

$\begin{array}{c}F=\left(4000\text{\hspace{0.17em}N/m}\right)(-0.200\text{\hspace{0.17em}m})\\ =-800\text{\hspace{0.17em}N}\end{array}$

 

Thus, the magnitude of force applied to hold the platform in position is.$800\text{\hspace{0.17em}N}$

Step 4: (b) Determination of theadditional work to be done to move the platform 0 .200  m farther and with maximum force

The total work done to move the platform is given by,

 

$W=\frac{1}{2}k{x}_2^2-\frac{1}{2}k{x}_1^2$

 

W is the force constant of the spring, $x_2$is the compressed length of the spring whose value is$0.400\text{\hspace{0.17em}m}$ ,$x_1$ is the initially compressed length.

 

Substitute all the values in the above,

$\begin{array}{c}W=\frac{1}{2}\left(4000\text{\hspace{0.17em}N/m}\right){\left(0.400\text{\hspace{0.17em}m}\right)}^2-\frac{1}{2}\left(4000\text{\hspace{0.17em}N/m}\right){\left(0\right)}^2\\ =320\text{\hspace{0.17em}J}\end{array}$

 

The additional work done required to move the platform is given by

$\begin{array}{c}{W}_{add}=W_{tot}-W\\ =(320-80)\text{\hspace{0.17em}J}\\ =240\text{\hspace{0.17em}J}\end{array}$,

 

The maximum force applied to the platform when it moves total distance$x=0.400\text{\hspace{0.17em}m}$ is given by,

 $F=kx$

 

Substitute all the values in the above,

 $\begin{array}{c}F=\left(4000\text{\hspace{0.17em}N/m}\right)\left(0.400\text{\hspace{0.17em}m}\right)\\ =1600\text{\hspace{0.17em}N}\end{array}$

 

Thus, the additional work done to move the platform and maximum force applied on the platform are$240\text{\hspace{0.17em}J}$ and$1600\text{\hspace{0.17em}N}$ respectively.