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Damm [24]
3 years ago
5

The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

e large mass, m2, falls through a height h and hits the floor, and the small mass, m2, rises through a height h.
Part A: Find the speed of the masses just before m2 lands, giving your answer in terms of m1, m2, g, and h. Assume the ropes and pulley have negligible mass and that friction can be ignored.
Part B: Evaluate your answer to part A for the case h = 1.7 m , m1 = 3.5 kg , and m2 = 4.3 kg.

Physics
1 answer:
Inga [223]3 years ago
6 0

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

E_i = E_f

0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

PART B) Replacing the values given as,

h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}

v_f = 1.8486m/s

Therefore the speed of the masses would be 1.8486m/s

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The resistance of a conductor is given by
R= \frac{\rho L}{A}
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We can see that if all the other quantities do not change, if the new length of the conductor is 4 times the original length: L'=4 L, then the new resistance is also 4 times the original value:
R'= \frac{\rho L'}{A}= \frac{\rho 4 L}{A}=4  \frac{\rho L}{A}=4 R
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In 1991 four English teenagers built an eletric car that could attain a speed of 30.0m/s. Suppose it takes 8.0s for this car to
ivanzaharov [21]

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6.3445×10⁻¹⁶ m

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E=\frac{mv^2}{2}\\\Rightarrow v=\sqrt{\frac{2E}{m}}\\\Rightarrow v=\sqrt{\frac{2\times 2470\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=29455356.08671\ m/s

Deflection by Earth's Gravity

\Delta =\frac {gt^2}{2}

Now, Time = Distance/Velocity

\Delta =\frac {g\frac{s^2}{v^2}}{2}\\\Rightarrow \Delta =\frac{9.81\frac{0.335^2}{29455356.08671^2}}{2}\\\Rightarrow \Delta =6.3445\times 10^{-16}\ m

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low pressure

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car starts from rest

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acceleration of the car will be

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now the time to reach this final speed will be

t = \frac{v_f - v_i}{a}

t = \frac{5 - 0}{3.6}

t = 1.39 s

so it required 1.39 s to reach this final speed

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