Answer:
a) potential energy in the spring = 1.8225 J
b) Jerry's speed = 2.744 m/s
Step-by-step explanation:
a) The spring constant, k = 1800 N/m
The rest length of the spring,
![L_(0) = 0.0550 m](https://img.qammunity.org/2021/formulas/physics/college/scedym6x7m7stfvchdaza93c977u4azcnv.png)
The launch length,
![L_(f) = 0.100 m](https://img.qammunity.org/2021/formulas/physics/college/hpp685flqbd61vabvk0lk7brsdw83d68cr.png)
The potential energy of each spring is ,
![PE = 0.5 k(\triangle x)^(2)](https://img.qammunity.org/2021/formulas/physics/college/4djo2ngt8wgjhizq0793527430lbr4by43.png)
![\triangle x = 0.1 - 0.055 = 0.045 m](https://img.qammunity.org/2021/formulas/physics/college/kz3pv81hgt03jx9wha718p30i0tck73t6t.png)
![PE = 0.5 * 1800* 0.045^(2)](https://img.qammunity.org/2021/formulas/physics/college/zbpij96xi3b3mk6gccyyifo3eyvu33m4mp.png)
PE = 1.8225 J
b) To get Jerry's speed, use the law of energy conservation
PE energy in the string = KE of the robot
![KE = 0.5mv^(2)](https://img.qammunity.org/2021/formulas/physics/college/uporou3vpkitjgsaplzczah4ryo0r4yqgs.png)
PE in the spring = 1.8225 J
![1.8825 = 0.5 mv^(2) \\1.8825 = 0.5 * 0.5 v^(2) \\1.8825 = 0.25 v^(2) \\ v^(2) = 1.8825/0.25\\v^(2) = 7.53](https://img.qammunity.org/2021/formulas/physics/college/c45ajwtej0dd2sxf919duygx3mrbhvvsjo.png)
v = 2.744 m/s