- The net force is greatest at the position of maximum displacement
- The net force is zero when at the equilibrium position
Explanation:
The motion of a spring is a Simple Harmonic Motion, in which the displacement of the end of the spring is given by a periodic function of the form

where A is the amplitude (the maximum displacement), and
the angular frequency of the motion.
We can analyze the net force acting on the spring by looking at Hooke's law:

where
F is the net force
k is the spring constant
x is the displacement
From the equation, we notice immediately that:
- The net force is the greatest when the displacement x is the greates, so at the position in which the spring has maximum compression or stretching
- The net force is zero when the displacement x is zero, so when the spring crosses the equilibrium position
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Answer:
The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1
Explanation:
Let the weight that is hooked to two springs be w.
Spring#1:
Force constant= k
let x1 be the extension in spring#1
Therefore by balancing the forces, we get
Spring force= weight
⇒k·x1=w
⇒x1=w/k
Energy stored in a spring is given by
where k is the force constant and x is the extension in spring.
Therefore Energy stored in spring#1 is, 
⇒
⇒
Spring #2:
Force constant= 2k
let x2 be the extension in spring#2
Therefore by balancing the forces, we get
Spring force= weight
⇒2k·x2=w
⇒x2=w/2k
Therefore Energy stored in spring#2 is, 
⇒
⇒
∴The ratio of the energy stored by spring #1 to that stored by spring #2 is
2:1
I think the puck pushes the stick backwards
Answer:
4 x 10⁻⁴ J
Explanation:
C = 5000 pF, V = 400 V
Energy = CV²/2 = 5000 x 10⁻¹² x 400²/2 = 4 x 10⁻⁴ J