A pan hangs from a 50 cm spring. When a 10 kg mass is placed in the pan, it stretches the spring 6 cm.

Question

#1  Write a function rule l(w) that models the length of the spring based on the mass of the object in the pan.

 

#2  What is the mass of an object in the pan if the kspring is stretched to a length of 65cm?

Answer 1

Use Hooke's Law

 

F_s = -kx,  where

 

F_s = the force exerted on the spring

k = spring constant

x = the elongation of the spring

 

Recall Newton's 2nd Law states that F = mg, so

 

Rewrite F_s as m_sg:

 

m_sg = -kx    where g is acceleration due to gravity.  g = 9.81 m/s²

 

Your rule is

-x = (m_sg)/k

 

The negative sign in front of x indicates a change in direction.  If you stretch a spring from one end, then the other end will move in the opposite direction due to the force exerted on it.  An example of this is using both of you hands to stretch a rubber band.  This is because all forces come in pairs based on Newton's 3rd Law:  F₁ = -F₂ .

 

For part 2, solve for m_s from the equation obtained in part 1.

 

Keep in mind the elongation is the different between the original length and the final length of the spring.