Question

A 150-g block on the end of a spring with a spring constant of 35N/m is pulled aside 25cm and released from rest. In the electrical analog the initial charge on the capacitor is:

Answer 1

Answer:

Q = 8.75 C.

Explanation:

In mechanics, when a spring is compressed, it stores potential energy. In electrostatics, when two plates are positioned closed and a potential difference is present, electrical potential energy is stored in the electric fields between the plates.

In the spring case, the initial potential energy is

$U_{elastic} = \frac{1}{2}kx^2 = \frac{1}{2}35(0.25)^2 = 1.09 ~J$

In the capacitor case, the potential energy is

$U = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{1}{2}\frac{Q^2}{C}$

The electrical analog of spring constant is capacitance. So,

$U = \frac{Q^2}{2C} = 1.09\\\frac{Q^2}{2(35)} = 1.09\\Q = 8.75~C$

Answer 2

In the electrical analog the initial charge on the capacitor is 8.75 C.

The given parameters;

• mass of the block, m = 150 g = 0.15 kg
• spring constant, k = 35 N/m
• extension of the spring, x = 25 cm = 0.25 m

The elastic potential energy of the spring is calculated as follows;

$E = \frac{1}{2} kx^2\\\\ E= \frac{1}{2} (35)(0.25)^2\\\\E = 1.094 \ J$

The energy stored in the capacitor is calculated as follows;

$E = \frac{Q^2}{2C}$

Where;

• C is the capacitance, let the capacitance = 35

$Q^2 = 2CE\\\\Q = \sqrt{2\times 35 \times 1.094}\\\\Q = 8.75 \ C$

Thus, in the electrical analog the initial charge on the capacitor is 8.75 C.

Learn more here:https://brainly.com/question/24943210