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SOVA2 [1]
3 years ago
15

A spring has a force constant of 500.0 N/m. Show that the potential energy stored in the spring is as follows: a. 0.400 J when t

he spring is stretched 4.00 cm from equilibrium b. 0.225 J when the spring is compressed 3.00 cm from equilibrium c. zero when the spring is unstretched
Physics
1 answer:
kogti [31]3 years ago
3 0

Answer:

(a) Hence, the potential energy = 0.400 J.

(b) Hence the potential Energy = 0.225 J.

(c) Hence the potential energy = 0 J

Explanation:

Potential Energy; This is the energy of a body, by virtue of its position in the gravitational field. The unit of potential energy is Joules (J)

the potential energy stored in a spring is

Ep = 1/2ke²..................................... Equation 1

Where Ep = potential Energy, k = force constant,  e = extension.

a.

When spring is stretched 4.00 cm,  e = 4.00 cm = 4/100 = 0.04 m, and k = 500 N/m.

Substituting into equation 1

Ep = 1/2(500)(0.04)²

Ep = 0.4 J.

Hence, the potential energy = 0.400 J.

(b)

When the spring is compressed 3.00 cm, e = 3.00 cm = 3/100 = 0.03 m, and k = 500 N/m.

Substitute into equation 1,

Ep = 1/2(500)(0.03)²

Ep = 250(0.0009)

Ep = 0.225 J.

Hence the potential Energy = 0.225 J.

(c) When the spring is unstretched, e = 0 cm = 0 m and k = 500 N/m.

Substituting into equation 1

Ep = 1/2(500)(0)²

Ep = 250(0)

Ep = 0.

Hence the potential energy = 0 J

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Two thin 80.0-cm rods are oriented at right angles to each other. Each rod has one end at the origin of the coordinates, and one
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Answer:

The net force on the electron is given as:

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Explanation:

Given:

charge on rod along x-axis = Q₁ = -15 x 10⁻⁶ C

charge on rod along y-axis = Q₂ = 15 x 10⁻⁶ C

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distance of electron from rod 1 = r₂ = 0.4 m

charge on electron = q = -1.6 x 10⁻¹⁹ C

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Electric force on charge due to rod 1:

F₁ = qE = 1/4πε°(qQ₁/r₁²)

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Negative negative repels each other so the rod will Force the electron in positive y-direction.

F₁ = 1.35 x 10⁻¹³ N j

Electric force on charge due to rod 2:

F₂ = qE = 1/4πε°(qQ₂/r₂²)

F₂ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x 15 x 10⁻⁶)/0.4²

F₂ = - 1.35 x 10⁻¹³ N

Opposite charges attract each other so the rod will force the electron in negative x-direction.

F₂ =  - 1.35 x 10⁻¹³ N i

Net Force:

F = F₁ + F₂

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

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