Respuesta :
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
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