Question

Block A is set on a rough horizontal table and is connected to a horizontal spring that is fixed to a wall, as shown.

Answer 1

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 = $\mathbf{\frac{1}{2} \cdot k \cdot x^2}$ > 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 = $\frac{1}{2} \cdot k \cdot x^2$

Energy received due to friction = 0

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

Where;

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

Which gives;

$m\cdot g \cdot h > m \cdot g \cdot (h - y) + \frac{1}{2} \cdot k \cdot x^2$

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;

brainly.com/question/5398294