Please help I’ve been stuck on this for a while!

Todd purchased a jumper for his infant son. The jumper clamps to the top of a door frame and has a spring from

which the body of the jumper hangs to allow the infant to jump up and down. Included with the jumper are small

weights, each identical in

weight and size. The weights allow the user to adjust the initial height of the jumper by

stretching the spring a certain distance. Todd adds 8 of the weights and notes the spring stretches 2 centimeters.

a. According to Hook's Law, the amount that a spring is stretched, s, varies directly with the force applied to the

spring by the weights, F. Write an equation that relates F and s for the spring on the jumper

b. Using your equation from part a, find the stretch in the spring if Todd applies 14 of the included weights to the

spring

c. Using your equation from part

Hooke's law states that provided the extension limit of an elastic material is not exceeded, the force applied is directly proportional to its extension.

i.e F = ke

where F is the force applied to the material, k is the spring constant of the material, and e is the extension.

Considering the conditions in the given question then, we have:

a. The equation that relates F and S for the spring on the jumper is:

F = ks

b. Given that: F = 8 N, while s = 2 cm (0.02 m).

Then;

F = ks

8 = k(0.02)

k = [tex]\frac{8.0}{0.02}[/tex]

= 400

k = 400 N/m

So that, the stretch in the spring when 14 N weight is included is:

F = ks

14 = 400 x s

s = [tex]\frac{14}{400}[/tex]

= 0.035

s = 0.035 m

Thus the spring was stretched by 0.035 m (3.5 cm) when a weight of 14 N was included.

Visit: https://brainly.com/question/20995298