work of 4 joules is done in stretching a spring from its natural length to 11 cm beyond its natural length. what is the force (in newtons) that holds the spring stretched at the same distance (11 cm)

The work performed on the spring is

W = 1/2 k x²

so that

4 J = 1/2 k (0.11 m)²   ⇒   k ≈ 660 N/m

Then by Hooke's law, the force required to hold the spring in this position is

F = k x = (660 N/m) (0.11 m) ≈ 73 N

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever