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leonid [27]
3 years ago
10

A 0.80 kg mass attached to an ideal spring oscillates horizontally with a period of 0.50 s and amplitude of 0.30 m. What is the

spring constant of the spring?
Physics
1 answer:
sasho [114]3 years ago
8 0

Answer:

126.33 N/m

Explanation:

Angular frquency is given by

f=\frac {2\pi}{t}

Wheret is period of wave and f is angular frequency.

Given period of 0.5 s then

w=\frac {2\pi}{0.5}=4\pi rad/s\approx 12.57 rad/s

Angular frequency is also given by

w=\sqrt{\frac {k}{m}}

Making k the subject then k=mw^{2}

Where k is spring constant, m is mass

Substituting m with 0.8 then

k=0.8\times (4\pi)^{2}

K=126.330936333944 N/m

Rounded off as 126.33 N/m

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Answer:

Part a)

10\hat i + 15\hat j = \vec v

Part b)

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m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

(2kg)(15\hat i + 30 \hat j) + (2 kg)(-10\hat i  + 5\hat j) = 2kg(-5\hat i + 20\hat j) + 2\vec v

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Part b)

magnitude of the initial speed of A = \sqrt{15^2 + 30^2} = 33.54 m/s

magnitude of the initial speed of B = \sqrt{10^2 + 5^2} = 11.18 m/s

magnitude of final speed of A = \sqrt{5^2 + 20^2} = 20.61 m/s

magnitude of final speed of B = \sqrt{10^2 + 15^2} = 18.03 m/s

Now change in total kinetic energy is given as

\Delta K = (\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2) - (\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2)

\Delta K = (\frac{1}{2}2(33.54)^2 + \frac{1}{2}2(11.18)^2) - (\frac{1}{2}2(20.61)^2 + \frac{1}{2}2(18.03)^2)

\Delta K = 500 J

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x=50

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