The left end of a spring is attached to a wall. When Bob pulls on the right end with a 200 N force, he stretches the spring by 20 cm. The same spring is then used for a tug-of-war between Bob and Carlos. Each pulls on his end of the spring with a 200 N force. How far does the spring stretch

Answer:

40 cm

Explanation:

Since Bob pulls with a force of F = 200 N when the spring is attached to the wall and it extends a length x = 20 cm, from Hooke's law,

F = kx where k = spring constant

So, k = F/x = 200 N/20 cm = 10 N/cm

Now, sine both Bob and Carlos pull with a force of F = 200 N in opposite directions, the spring stretches about its center and has an extension, x' in each direction.

So, from F = kx

x = F/k = 200 N/10 N/cm = 20 cm

So, the spring stretches 20 cm in both directions.

So, the total extension is thus x' + x' = 2x' = 2(20 cm) = 40 cm

The spring will stretch 40 cm.When a force is act om the spring the spring will stretch or compress depeds on the application of force.

Hooke's law states that the force used to extend the spring is directly equal to the amount of stretch.

The force needed to extend the spring is proportional to its displacement. It is stated as

F=Kx

The given data in the problem is;

F is the force of pull to bob=  200 N

x is the length of extension= 20 cm.

F' is the force act on the another end=  200 N

According to Hooke's law,

[tex]\rm F = Kx \\\\ \rm K= \frac{F}{x} \\\\ \rm K= \frac{200}{20} \\\\ \rm K=10 \ N/cm[/tex]

The extension due to another end force is found by;

[tex]\rm x' = \frac{F}{K} \\\\ \rm x' = \frac{200}{10} \\\\ \rm x' = 20[/tex]

The total extension of the spring will be;

[tex]\rm x_t = x+x' \\\\ \rm x_t = 2(20) \\\\ \rm x_t =40\ cm[/tex]

Hence the spring will stretch 40 cm.

To learn more about the hooke's law refer to the link;

https://brainly.com/question/13348278

The _____________ variable is observed, measured, and affected by the independent variable.​