91Ó°ÊÓ

The given data can be listed below,

• The work done on the spring to compress it is,$W=80\text{ J}$
• The compressed length of the spring is,$x_2=0.200\text{″¾}$

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.

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,

$\begin{array}{c}k=\frac{2×80\text{ J}}{{\left(0.200\text{″¾}\right)}^2−{\left(0\right)}^2}\\ =4000\text{ N/³¾}\end{array}$

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,

$\begin{array}{c}F=\left(4000\text{ N/³¾}\right)(−0.200\text{″¾})\\ =−800\text{ N}\end{array}$

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

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$0.400\text{″¾}$ ,$x_1$ is the initially compressed length.

Substitute all the values in the above,

$\begin{array}{c}W=\frac{1}{2}\left(4000\text{ N/³¾}\right){\left(0.400\text{″¾}\right)}^2−\frac{1}{2}\left(4000\text{ N/³¾}\right){\left(0\right)}^2\\ =320\text{ J}\end{array}$

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

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

The maximum force applied to the platform when it moves total distance$x=0.400\text{″¾}$ is given by,

$F=kx$

Substitute all the values in the above,

$\begin{array}{c}F=\left(4000\text{ N/³¾}\right)\left(0.400\text{″¾}\right)\\ =1600\text{ N}\end{array}$

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