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tekilochka [14]
3 years ago
14

A 14.6 g mass is attached to a horizontal spring with a spring constant of 15.7 N/m and released from rest with an amplitude of

29.8 cm. What is the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless
Physics
1 answer:
Tanya [424]3 years ago
7 0

Answer:

v = 8.46 m/s

Explanation:

Given that,

Mass of the object, m = 14.6 g

Spring constant of the spring, k = 15.7 N/m

It is released from rest with an amplitude of 29.8 cm.

We need to find the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless.

We can use the conservation of energy in this case. Let v be the speed of the mass. So,

EPE_i=\dfrac{1}{2}mv^2+EPE_f\\\\\dfrac{1}{2}kx_i^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx_f^2

Substitue all the values in the above expression.

\dfrac{1}{2}\times 15.7\times (0.298)^2=\dfrac{1}{2}\times 0.0146\times v^2+\dfrac{1}{2}\times 15.7\times (\dfrac{0.298}{2})^2\\\\v=8.46\ m/s

So, the speed of the mass is equal to 8.46 m/s.

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Objects with masses of 255 kg and a 555 kg are separated by 0.390 m.
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Answer:

(a). The net gravitational force is 4.20\times10^{-6}\ N

(b). The position is at 0.232 m.

Explanation:

Given that,

Mass of one object M = 255 kg

Mass of another object M'= 555 kg

Separation = 0.390 m

(a). We need to calculate the net gravitational force

Using formula of force

F_{net}=\dfrac{GM'm}{r^2}-\dfrac{GMm}{r^2}

F_{net}=\dfrac{Gm(M'-M)}{r^2}

Put the value into the formula

F_{net}=\dfrac{6.67\times10^{-11}\times32.0(555-255)}{(0.390)^2}

F_{net}=4.20\times10^{-6}\ N

The net gravitational force is 4.20\times10^{-6}\ N

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Force from 555 mass = Force from 255 mass

\dfrac{Gm\times555}{x^2}=\dfrac{Gm\times255}{(0.390-x)^2}

555(0.390-x)^2=255x^2

300x^2-0.78x+84.4155=0

x=0.232\ m

The position is at 0.232 m.

Hence, (a). The net gravitational force is 4.20\times10^{-6}\ N

(b). The position is at 0.232 m.

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