A spring that has a spring constant of 1400 N/m is stretched to a length of 2.5 m. If the normal length of the spring is 1.0 m,
2 answers:
Answer:
1575 J
Explanation:
The elastic potential energy of a string is given by:
where
k is the spring constant
is the stretching/compression of the string with respect to its natural length
For the spring in the problem, we have:
- Spring constant:
- Stretching:
Therefore, the elastic potential energy stored in the spring is
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Answer:
Explanation:
plate separation = 2.3 x 10⁻³ m
capacity C₁ = ε A / d
= ε A / 2.3 x 10⁻³
C₂ = ε A / 1.15 x 10⁻³
=
a ) when charge remains constant
energy =
q is charge and C is capacity
energy stored initially E₁=
energy stored finally E₂ =
=
=
=
= 4.19 J
b )
In this case potential diff remains constant
energy of capacitor = 1/2 C V²
energy is proportional to capacity as V is constant .
= 16.76 .