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3 years ago

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A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10

cm/s and no damping is applied.
(a) Determine the position u of the mass at any time t. Use 9.8 m/s as the acceleration due to gravity. Pay close attention to the units.
(b) When does the mass first return to its equilibrium position?

1 answer:

Charra [1.4K]3 years ago

6 0

Answer:

$u(t)=1.15 \sin (8.68t)cm$

0.3619sec

Explanation:

Given that

Mass,m=148 g

Length,L=13 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

$\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s$

Where $g=980 cm/s^2$

$u(t)=Acos8.68 t+Bsin 8.68t$

u(0)=0

Substitute the value

$A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t$

Substitute u'(0)=10

$8.68B=10$

$B=\frac{10}{8.68}=1.15$

Substitute the values

$u(t)=1.15 \sin (8.68t)cm$

Period =T = 2π/8.68

After half period

π/8.68 it returns to equilibruim

π/8.68 = 0.3619sec

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Answer:

a) $c=1822.3214\ J.kg^{-1}.K^{-1}$

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c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

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Given:

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mass of water, $m_w=0.25\ kg$

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temperature of copper block, $T_c=50^{\circ}C$

mass of the other block, $m=0.07\ kg$

temperature of the other block, $T=100^{\circ}C$

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specific heat of copper, $c_c=390\ J.kg^{-1}.K^{-1}$

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The heat absorbed by the system of the calorie-meter to reach the final temperature.

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The heat released by the blocks when dipped into water:

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This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c)

The material is peat, possibly.

d)

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

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