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lana [24]
2 years ago
14

The Atwood’s machine shown consists of two blocks of mass m1 and m2 that are connected by a light string that passes over a pull

ey of negligible friction and negligible mass. The block of mass m1 is a distance h1 above the ground, and the block of mass m2 is a distance h2 above the ground. m2 is larger than m1. The two-block system is released from rest. Which of the following claims correctly describes the outcome after the blocks are released from rest but before the block of m2 reaches the ground? Select two answers.

Physics
1 answer:
Talja [164]2 years ago
5 0

(A) For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

(D) For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.

<h3 /><h3>The given parameters:</h3>
  • Mass of block 1 = m1
  • Mass of block 2, = m2
  • Height of block 1 above the ground, = h1
  • Height of block 2 above the ground = h2

The total initial mechanical energy of the two block system is calculated as follows;

m_1gh_1 + \frac{1}{2} m_1v_1_i^2 = m_2gh_2 + \frac{1}{2} m_2v_2_i^2\\\\m_1gh_1 + 0 = m_2gh_2 + 0\\\\m_1gh_1 = m_2gh_2\\\\m_1gh_1 - m_2gh_2 = 0

When the block m2 reaches the ground the block m1 attains maximum height and the total mechanical energy at this point is given as;

m_1g(h_1 + h_2) + K.E_1 = \frac{1}{2}m_2v_{max}^2 + P.E_2\\\\m_1g(h_1 + h_2 ) -PE_2 = \frac{1}{2}m_2v_{max}^2 - K.E_1\\\\m_1g(h_1 + h_2 )  - 0= \frac{1}{2}m_2v_{max}^2 - 0\\\\m_1g(h_1 + h_2 )  = \frac{1}{2}m_2v_{max}^2\\\\W = \frac{1}{2}m_2v_{max}^2

Thus, we can conclude the following before the block m2 reaches the ground;

  • For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
  • For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

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It took 2006 to 2015 (9 years) to travel from Earth to Pluto, therefore, it can take approximately (9 years)/(23.683) ≈ 0.38 of a year which is ((9 years)/(23.683)) × 365.2422 ≈ 138.8 days for a spacecraft to travel from Earth to Mars

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What is the sequence of energy transformations from the moment the ball is dropped to the moment the board is bent to its maximu
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(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

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where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

B=\rho_a V g

where \rho_a is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

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The buoyant force is given by

B=\rho_a V g

where \rho_a is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

\rho_a = 1.29 kg/m^3 is the air density

V=329 m^3 is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N

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The mass of the helium gas inside the balloon is

m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg

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m=m_h+m_b=59 kg+210 kg=269 kg

So now we can find the weight of the balloon:

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F=B-W=4159 N-2635 N=1524 N

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(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

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(f) The balloon and its load will accelerate upward.

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