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vivado [14]
3 years ago
7

How does the work needed to lift an object and the gravitational potential energy of an object compare

Physics
1 answer:
Ganezh [65]3 years ago
8 0

When you do work to lift the object, the amount of work you do BECOMES the object's gravitational potential energy.  It GETS its potential energy from the work you do to lift it.  They're equal.  You lose it, and the object gains it.  Energy is not created or destroyed.  It's just transferred from you to the object.  

Later, when you DROP the object, GRAVITY does the same amount of work on it, to pull it to the ground.  Again, no energy is created or destroyed.  Every time a force acts to move anything, the energy to do it comes from somewhere, and the energy goes somewhere.

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A uniform, 4.5 kg, square, solid wooden gate 2.0 mm on each side hangs vertically from a frictionless pivot at the center of its
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Answer:

The angular velocity is  w = 1.43\  rad/sec

Explanation:

From the question we are told that

   The  mass of wooden gate  is m_g = 4.5 kg

    The  length of side is  L = 2 m

    The mass of the raven is  m_r = 1.2 kg

     The initial speed of the raven is u_r = 5.0m/s

     The final speed of the raven is   v_r = 1.5 m/s

From the law of  conservation of angular momentum we express this question mathematically as

       Total initial angular momentum  of both the Raven and  the Gate =  The Final angular momentum of both the Raven and the Gate  

The initial angular momentum of the Raven is m_r * u_r * \frac{L}{2}

Note: the length is half because the Raven hit the gate at the mid point

The initial angular momentum of the Gate is  zero

Note: This above is the generally formula for angular momentum of  square objects

  The final angular velocity  of the Raven is  m_r * v_r * \frac{L}{2}

   The  final angular velocity of the Gate  is   \frac{1}{3} m_g L^2 w

Substituting this formula

  m_r * u_r * \frac{L}{2}  =   \frac{1}{3} m_g L^2 w + m_r * v_r * \frac{L}{2}

  \frac{1}{3} m_g L^2 w   =    m_r * v_r * \frac{L}{2} -   m_r * u_r * \frac{L}{2}

  \frac{1}{3} m_g L^2 w   =    m_r *  \frac{L}{2} * [u_r - v_r]

Where w is the angular velocity

     Substituting value  

   \frac{1}{3} (4.5)(2)^2  w   =    1.2 *  \frac{2}{2} * [5 - 1.5]

     6w = 4.2

       w = \frac{6}{4.2}

            w = 1.43\  rad/sec

5 0
3 years ago
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