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kykrilka [37]
3 years ago
10

In the system drawn on the right, m_1 = 2.0 kg sits on a lab table which is 110 cm tall. m_2 = 1.0 kg is hanging over a pulley a

nd is initially at rest 60.0 cm above the floor. The system is released. Determine the speed of m_2 just before it hits the floor. (solve with work-energy ideas, considering the system)
a) solve the problem assuming the table is frictionless.
b) coefficient of friction is .2 between m1 and the table
Physics
1 answer:
laila [671]3 years ago
8 0

(a) The speed of the mass m2 just before it hits the floor for frictionless table is 1.98 m/s.

(b) The speed of the mass m2 just before it hits the floor when there is friction is 1.53 m/s.

<h3>Apply Newton's second law of motion for object m₁</h3>

-f_k + T = m_1a\\\\&#10;T = m_1 a + f_k \ ---\ (1)\\\\&#10;

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

m_2g - T = m_2a \ ---- (2)\\\\&#10;

Solve (1) and (2) together

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

<h3>Speed of the mass for frictionless table</h3>

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

<h3>The speed of the mass when there is friction</h3>

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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