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Block A is set on a rough horizontal table and is connected to a horizontal spring that is fixed to a wall, as shown.
Block A is then also connected to hanging block B by a lightweight string that passes over an ideal pulley, as
shown. The friction force exerted on block A by the table is not negligible. The blocks are initially held at rest so
that the spring is not stretched. When the blocks are released, hanging block B moves downward and block A on
the table moves to the right until the system comes again to rest. Let E1 be the mechanical energy of the blocks-
spring system, and let E2 be the mechanical energy of the blocks-spring-Earth system. How do these two energies
change from when the blocks are held at rest to when the blocks come to rest again?

Sagot :

The mechanical energy of the system is sum of the potential and kinetic

energy of the system.

Response:

  • E1 increases, E2 decreases

Methods by which the the above response is obtained

Mechanical energy, M.E. in the block and spring system can be presented as follows;

M.E. = Energy in the spring + Kinetic energy of the blocks + Energy done on friction

  • Mechanical energy of the block-spring system, E1

When the blocks are held at rest;

The mechanical energy in the block-spring system when the blocks are held at rest can be found as follows;

Energy in the spring = 0

Kinetic energy of the blocks = 0

Friction energy = 0

Therefore;

E1 for the block at rest = 0

E1 when the blocks come to rest again

Energy in the spring = [tex]\mathbf{\frac{1}{2} \cdot k \cdot x^2}[/tex] > 0

Kinetic energy = 0

Energy of friction = 0

Therefore;

The mechanical energy of the block-spring system, E1, increases

  • The mechanical energy of the block-spring-Earth system, E2

When the blocks are held

Energy in the spring = 0

Energy done due to friction = 0

Potential energy of Block B = m·g·h

Kinetic energy of the blocks = 0

When the blocks come to rest again, we have;

Energy in the spring = [tex]\frac{1}{2} \cdot k \cdot x^2[/tex]

Energy received due to friction = 0

Potential energy of Block B = m·g·(h - y)

Where;

[tex]m \cdot g \cdot h = \mathbf{m \cdot g \cdot (h - y) + \frac{1}{2} \cdot k \cdot x^2 + Energy \ loss \ due \ to \ friction}[/tex]

Which gives;

[tex]m\cdot g \cdot h > m \cdot g \cdot (h - y) + \frac{1}{2} \cdot k \cdot x^2[/tex]

The energy in the spring-block-Earth system, E2, when initially held is more than the the energy when the blocks com to rest again.

Therefore, E2 decreases

The correct option is therefore;

  • E1 increases, E2 decreases

The possible question options obtained from a similar question posted online are;

  • E1 increases, E2 decreases
  • E1 decreases, E2 increases
  • E1 is constant, and E2 decrease
  • E1 and E2 remain constant

Learn more about mechanical energy here;

https://brainly.com/question/5398294

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