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Alecsey [184]
2 years ago
9

Mass A forklift raises a box 1.2 m and does 7.0 kJ of work on it. What is the mass of the box?

Physics
1 answer:
Mamont248 [21]2 years ago
5 0

Answer:

Work and Kinetic Energy

A B

3. A 0.180 kg balls falls 2.5 m. How much work does the force of gravity do on the ball? 4.41 J

4. A forklift raises a box 1.2 m doing 7.0 kJ of work on it. What is the mass of the box? 595.24 kg

5. How much work does the force of gravity do when a 25 N object falls a distance of 3.5 m? 87.5 J

Explanation:

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In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s
riadik2000 [5.3K]

Answer:

Part a)

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

\tau = I\alpha

here we will have

\tau = (m_1g - m_2g)(\frac{L}{2})

now we will have inertia of two masses given as

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now we have

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now the angular acceleration is given as

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

so we have

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

Now if the rod is not massles then we will have total inertia given as

I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}

so we will have

I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}

now the acceleration is given as

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

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

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3 years ago
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