A person wants to determine the spring constant of an exercise stretch cord. He pulls the cord with a force probe that exerts a 35-N force on the cord, causing it to stretch 20 cm. Using the Hooke's law find the spring constant of the cord. Assume that the stretch does not exceed the elastic range of the material. Express your answer to two significant figures and include the appropriate units.

180N/m(to 2 significant figures)

Explanation:

According to hooked law which states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically, F = ke where;

F is the applied force in newtons

k is the elastic/spring constant in N/m

e is the extension in meters

Given applied force = 35N

extension = 20cm = 0.2m

Since F = ke,

k = F/e = 35/0.2

k = 175N/m

The spring constant is 175N/m

= 180N/m (to 2significant figures)

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft