91Ó°ÊÓ

The given data can be listed below,

• The mass of the performer is $m=60\text{ k²µ}$,
• The force constant of the spring is role="math" $k=1100\text{ N/³¾}$,
• The restoring force on the spring is $F_{ext}=4400\text{ N}$,
• The average frictional force is $f_k=40\text{ N}$,
• The final distance travelled is $y=2.5\text{″¾}$,

The energy an item or body possesses as a result of its motion is known as kinetic energy. It often depends on the frame.

The initial kinetic and gravitational energy of the performer is zero as he is initially at rest

The initial compressed length of the spring is given by restoring force,

$\begin{array}{l}F=kd\\ d=\frac{F}{k}\end{array}$

Here, F is the restoring force and k is the force constant

Substitute all the values in the above,

$\begin{array}{c}d=\frac{4400\text{ N}}{1100\text{ N/³¾}}\\ =4.0\text{″¾}\end{array}$

The elastic potential energy is given by,

$U_{el}=\frac{1}{2}k{d}^2$

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

Substitute all the values in the above,

$\begin{array}{c}{U}_{el}=\frac{1}{2}\left(4400\text{ N/³¾}\right){\left(4.0\text{″¾}\right)}^2\\ =8800\text{ J}\end{array}$

The final gravitational potential energy is given by,

$U_{grav}=mgy$

Here, m is the mass of the performer, g is the acceleration due to gravity, and y is the distance

travelled above the initial position.

Substitute all the values in the above,

$\begin{array}{c}{U}_{grav,2}=\left(60\text{ k²µ}\right)\left(9.8{\text{″¾/²õ}}^2\right)\left(2.5\text{″¾}\right)\\ =15\left(9.8{\text{″¾/²õ}}^2\right)\\ =1470\text{ J}\end{array}$

The other force acting on the system is given by,

$F_{other}=−fs$

Substitute all the values in the above,

$\begin{array}{c}{F}_{other}=−\left(40\text{ N}\right)\left(4.0\text{″¾}\right)\\ =−160\text{ J}\end{array}$

The final velocity of the performer is given by law of conservation of energy.

$\begin{array}{c}{K}_1+U_{grav,1}+F_{other}=K_2+U_{grav,2}\\ 0+8800\text{ J}−160\text{ J}=\frac{1}{2}m{v}_2^2+1470\text{ J}\\ v_2^2=\frac{2×7170\text{ J}}{m}\end{array}$

Substitute all the values in the above,

$\begin{array}{c}{v}_2=\sqrt{\frac{2×7170\text{ J}}{m}}\\ =\sqrt{\frac{14340\text{ J}}{60\text{ k²µ}}}\\ =\sqrt{239{\text{″¾}}^{\text{2}}{\text{/s}}^{\text{2}}}\\ =15.5\text{″¾/²õ}\end{array}$

Thus, the velocity emerged by the performer is $15.5\text{″¾/²õ}$.