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Delvig [45]
1 year ago
11

Help Please!Could someone help me out with this one?

Physics
1 answer:
Radda [10]1 year ago
5 0

mass of the chair = 25 kg

Force required to start the motion = 165 N

Force required to maintain the motion = 127 N

part a)

when we applied the force to just start the motion then it is just equal to the limiting friction force

F_s = F

now as we know that

\mu_s mg = 165

\mu_s * 25 * 9.8 = 165

\mu_s = \frac{165}{25*9.8}

\mu_s = 0.67

PART b)

when object has started motion then to maintain its motion we need external force to balance kinetic friction  

F_k = F

now as we know that

\mu_k mg = 127

\mu_k * 25 * 9.8 = 127

\mu_k = \frac{127}{25*9.8}

\mu_k = 0.52

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<h3>Apply Newton's second law of motion for object m₂</h3>

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m_2 g - (m_1 a+ f_k) = m_2 a\\\\&#10;m_2 g - f_k = m_1 a + m_2 a\\\\&#10; m_2 g - f_k = a (m_1 + m_2)\\\\&#10;a = \frac{m_2 g- f_k}{m_1 + m_2}

When the table is frictionless, the acceleration of the masses is calculated as follows;

a = \frac{m_2 g}{m_1 + m_2} \\\\&#10;a = \frac{1 \times 9.8}{2 + 1}  \\\\&#10;a = 3.27 \ m/s^2

When the coefficient of friction between m1 and the table is 0.2, the acceleration of the masses is calculated as follows;

a = \frac{m_2 g - f_k}{m_ 1+ m_2} \\\\&#10;a = \frac{m_2 g - \mu_k m_1g}{m_1 + m_2} \\\\&#10;a = \frac{g(m_2 - \mu_km_1)}{m_1 + m_2} \\\\&#10;a = \frac{9.8(1 - \ 0.2\times 2)}{2+1 } \\\\&#10;a = 1.96 \ m/s^2

The height of the mass m2 above the ground = 60 cm = 0.6

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The speed of the mass m2 just before it hits the floor for frictionless table is calculated as follows;

v^2 = u^2 + 2ah\\\\&#10;v^2 = 0 + 2ah\\\\&#10;v^2 = 2ah\\\\&#10;v= \sqrt{2ah} \\\\&#10;v = \sqrt{2 \times 3.27  \times 0.6} \\\\&#10;v = 1.98 \ m/s

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The speed of the mass m2 just before it hits the floor when there is friction is calculated as follows;

v = \sqrt{2ah} \\\\&#10;v = \sqrt{2 \times 1.96 \times 0.6} \\\\&#10;v = 1.53 \ m/s

Learn more about acceleration mass over a pulley here: brainly.com/question/13376070

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3 points
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A and D

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