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3 years ago

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A 0.15 kg mass is suspended from a vertical spring and descends a distance of 4.6 cm, after which it hangs at rest. An additiona

l 0.50 kg mass is then suspended from the first. What is the total extension of the spring? (Neglect the mass of the spring.)a. FOLLOW UP: How much work is done by gravity in stretching the spring through both displacements?

1 answer:

ch4aika [34]3 years ago

5 0

Answer:

The total extension = 19.9 cm

Work done in stretching the spring through both displacement = <em>0.633 J.</em>

Explanation:

From Hook's law,

F = ke .................................... Equation 1.

Where F = force or weight, k = force constant of the spring, e = extension.

making k the subject of the equation,

k = F/e .......................... Equation 2

Given: F = W = mg = 0.15(9.8) = 1.47 N, e = 4.6 cm = 0.046 m.

k = 1.47/0.046

k = 31.96 N/m.

When an additional mass of 0.5 kg is suspended,

Total mass suspended(M₁) = 0.15+0.5 = 0.65 kg

Total Weight (W₁) = 0.65(9.8) = 6.37 N.

k = 31.96 N/m

Substituting into equation 1

6.37 = 31.96e

e = 6.37/31.96

e = 0.199 m

e = 19.9 cm

Thus the total extension = 19.9 cm

a.

Work done in stretching the spring through both displacement

W = 1/2ke²..................... Equation 3

Where W = work done, k = spring constant, e = extension.

Given: k = 31.96 N/m, e = 0.199 m

Substituting into equation 3

W = 1/2(31.96)(0.199)²

<em>W = 0.633 J.</em>

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Part b)

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Part c)

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Part 4)

angular acceleration will be ZERO

Part 5)

$I = 345.6 kg m^2$

Explanation:

Part a)

Moment of inertia of the system about left end of the rod is given as

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$I = \frac{m_r(4R)^2}{3} + (\frac{2}{5}(5m_r) R^2 + (5m_r)(R + 4R)^2)$

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$I = (\frac{16}{3} + 127)m_r R^2$

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Part b)

If force is applied to the mid point of the rod

so the torque on the rod is given as

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now angular acceleration is given as

$\alpha = \frac{\tau}{I}$

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$\alpha = 0.70 rad/s^2$

Part c)

position of center of mass of rod and sphere is given from the center of the sphere as

$x = \frac{m_r}{m_r + m_s}(\frac{L}{2} + R)$

$x = \frac{m_r}{6 m_r}(3R) = \frac{R}{2}$

so moment of inertia about this position is given as

$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + \frac{R}{2})^2 + (\frac{2}{5} m_s R^2 + m_s(\frac{R}{2})^2)$

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$I = \frac{m_r (16R^2)}{12} + m_r(\frac{5R}{2})^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(\frac{R^2}{4})$

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Part 4)

If force is applied parallel to the length of rod

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Part 5)

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$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + 2R)^2 + (\frac{2}{5} m_s R^2 + m_s(R)^2)$

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$I = 345.6 kg m^2$

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3 years ago