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lbvjy [14]
3 years ago
9

Change in Energy Quick Check

Physics
1 answer:
nika2105 [10]3 years ago
4 0

Answer:

The force of gravity on the object decreases.  (FALSE)

The potential energy of the object decreases.  (TRUE)

The acceleration due to gravity decreases.  (FALSE)

The kinetic energy of the object decrease (FALSE)

Explanation:

<u>FORCE OF GRAVITY:</u>

Force of gravity on the object is the weight of object, which depends upon the mass and value of acceleration due to gravity (W = mg). Since, the value mass is constant on Earth and acceleration due to gravity is also constant on earth.

Therefore, force of gravity remains same on the object. <u>The statement is false.</u>

<u></u>

<u>POTENTIAL ENERGY:</u>

Potential energy of object depends upon the mass, value of acceleration due to gravity, and change in height of object (P.E = mgΔh). Since, the value mass is constant on Earth and acceleration due to gravity increases as the object moves towards the surface of earth. But, the height of object is decreasing.

Therefore, potential energy of object decreases. <u>The statement is true.</u>

<u></u>

<u>ACCELERATION DUE TO GRAVITY:</u>

Acceleration due to gravity depends upon the altitude (gh = g[1 - 2h/Re]). Since, the height of object is decreasing.

Therefore, acceleration due to gravity increases. <u>The statement is false.</u>

But, this change is not significant.

<u>KINETIC ENERGY:</u>

Kinetic Energy of object, which depends upon the mass and velocity of the object (K.E = mv²/2). Since, the value mass is constant on Earth and velocity increases as the object moves towards the surface of earth.

Therefore, kinetic energy of the object also increases. <u>The statement is false.</u>

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Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

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Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

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