Suppose a force of 60 n is required to stretch and hold a spring 0.1 m from its equilibrium position. a. assuming the spring obeys hooke's law, find the spring constant k. b. how much work is required to compress the spring 0.5 m from its equilibrium position? c. how much work is required to stretch the spring 0.6 m from its equilibrium position? d. how much additional work is required to stretch the spring 0.1 m if it has already been stretched 0.1 m from its equilibrium? a. kequals 600

a. 600 N/m

Hooke's law states that:

F = kx

where

F is the force applied

k is the spring constant

x is the stretching/compression of the spring relative to the equilibrium position

In this problem we have

F = 60 N

x = 0.1 m

So the spring constant is

b. 75 J

The work required to stretch a spring is equal to the elastic potential energy stored in the spring:

where

k is the spring constant

x is the stretching/compression of the spring

Here we have

k = 600 N/m

x = 0.5 m

So the work done is

c. 108 J

We can use the same formula used in the previous part:

where here we have

k = 600 N/m

x = 0.6 m

So the work done is

d. 9 J

In this case, the additional work required is the difference between the elastic potential energy in the two situations

where

k = 600 N/m

is the initial stretching

is the final stretching

Solving the equation,

dang that balls went far