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Leona [35]
3 years ago
15

A 1.5-kg mass attached to an ideal massless spring with a spring constant of 20.0 N/m oscillates on a horizontal, frictionless t

rack. At time t = 0.00 s, the mass is released from rest at x = 10.0 cm. (That is, the spring is stretched by 10.0 cm.) (a) Find the frequency of the oscillations and express the displacement x as a function of time t. (b) Determine the maximum speed and the maximum acceleration of the mass. (c) At what location are the kinetic energy and the potential energy of the system the same?
Physics
1 answer:
Molodets [167]3 years ago
6 0

Answer:

Explanation:

a ) angular frequency ω = \sqrt{\frac{k}{m} }

k is spring constant and m is mass attached

ω = \sqrt{\frac{20}{1.5} }

= 3.6515 rad / s

frequency of oscillation n = 3.6515 / (2 x 3.14)

= .5814 s⁻¹

x = .1 mcos(ωt)

= .1 mcos(3.6515t)

b ) maximum speed = ωA , A is amplitude

= 3.6515 x .1

= .36515 m /s

36.515 cm /s

maximum acceleration = ω²A

= 3.6515² x .1

= 1.333 m / s²

c ) Kinetic energy at displacement x

= 1/2 m ω²( A²-x²)

potential energy =1/2 m ω²x²

so 1/2 m ω²( A²-x²) = 1/2 m ω²x²

A²-x² = x²

2x² = A²

x = A / √2

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3 years ago
From the top of the engineering building, you throw a ball vertically upward. the ball strikes the ground 4.00 s later. the engi
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Equations of the vertical launch:

Vf = Vo - gt

y = yo + Vo*t - gt^2 / 2

Here yo = 35.0m
Vo is unknown
y final = 0
t = 4.00 s
and I will approximate g to 10m/s^2

=> 0 = 35.0 + Vo * 4 - 5 * (4.00)^2 => Vo = [-35 + 5*16] / 4 = - 45 / 4 = -11.25 m/s

The negative sign is due to the fact that the initial velocity is upwards and we assumed that the direction downwards was positive when used g = 10m/s^2.

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A heavy hall with a mass of 3.5 kg is observed to accelerate
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Answer:

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Explanation:

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force = mass × acceleration

From the question we have

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We have the final answer as

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Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (
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Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

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4 kg (2.5, 0)

8.3 kg (x, y)

As we know that the formula of center of gravity is given as

x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}

10 + 8.3 x = 0

x = -1.20 m

Similarly for y direction we have

y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}

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3 years ago
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