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vagabundo [1.1K]
4 years ago
14

A 50-cm-long spring is suspended from the ceiling. A 270 g mass is connected to the end and held at rest with the spring unstret

ched. The mass is released and falls, stretching the spring by 24 cm before coming to rest at its lowest point. It then continues to oscillate vertically. Part A What is the spring constant?
Physics
1 answer:
AveGali [126]4 years ago
5 0

Answer:

k = 22.05 N/m

Explanation:

The potential energy of the mass is converted into potential energy of the spring.

Given:

mass m = 0.27 kg

gravitational constant g = 9.8 m/s²

distance falling/ stretching of spring h = 0.24 m

U_{gravity} = U_{spring}\\ mgh = \frac{1}{2} kh^{2}

Solving for k:

k = 2mg\frac{1}{h}

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Eddi Din [679]

Answer:

\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }

Explanation:

Additional information:

<em>The ball has charge </em>-q_0<em>, and the ring has  positive charge </em>+Q<em> distributed uniformly along its circumference. </em>

The electric field at distance z along the z-axis due to the charged ring is

E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.

Therefore, the force on the ball with charge -q_0 is

F=-q_oE_z

F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}

and according to Newton's second law

F=ma=m\dfrac{d^2z}{dz^2}

substituting F we get:

- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}

rearranging we get:

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0

Now we use the approximation that

z^2+a^2\approx a^2 <em>(we use this approximation instead of the original </em>d^2+a^2\approx a^2<em> since </em>z<em>, our assumption still holds )</em>

and get

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0

Now the last equation looks like a Simple Harmonic Equation

m\dfrac{d^2z}{dz^2}+kz=0

where

\omega=\sqrt{ \dfrac{k}{m} }

is the frequency of oscillation. Applying this to our equation we get:

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}

\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}

5 0
4 years ago
4. What two terms are sometimes used to indicate opposite directions from a point?
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<h2><em>What two terms are sometimes used to indicate opposite directions from a point?</em></h2>

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The 0.01 kg marble is dropped from rest at A through the smooth glass tube and accumulate in the basket at C as shown in Figure
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denis23 [38]

Answer:

2 revolutions

Explanation:

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s = gt^2/2

where s = 8.3 m is the distance that she falls, t is the time it takes to fall, which is what we are looking for

t^2 = \frac{2s}{g} = \frac{2*8.3}{9.8} = 1.694

t = \sqrt{1.694} = 1.3 s

Since she rotates with an average angular speed of 1.6rev/s. The number of revolutions she would make within 1.3s is

rev = 1.3 * 1.6 = 2 revolution

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