Question

If the cylinder is originally not rotating and the block is released from rest at a height 1.54 m above the ground, to what height will this mass rebound if it bounces straight back up from the floor

Answer 1

If this mass bounces straight back up from the floor, it will rebound at a height of 1.68 m.

What is the law of conservation of energy?

According to the Law of Conservation of Energy, energy can neither be created nor destroyed, but it can be transferred from one form to another.

The total energy is the sum of all the energies present in the system. The potential in a system is due to its position in the system.

energy

From the law of conservation of energy principle;

$\rm \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh \\\\ \frac{1}{2} mv^2 +\frac{1}{2}mr^2 \times (\frac{v}{r})^2=mgh$

Substitute the given values;

$\rm \frac{1}{2} (1.68)v^2+0.25(0.56)(v^2)= 1.68 \times (9.8)(1.54) \\\\ v= 5.72 \ m/sec \\\\ 0.84 v^2+0.14 v^2= 25.35 \\\\ 0.98 v^2 = 25.35 \\\\ v^2=25.86 \\\\ v= 5.08 \ m/sec$

The height up to which the masses rebound;

$\rm h' = \frac{v^2}{2g} \\\\\\ h'=\frac{5.72^2}{2 (9.81)} \\\\ h'=1.67 \ m$

Hence, if this mass bounces straight back up from the floor, it will rebound at a height of 1.68 m.

To learn more about the law of conservation of energy, refer to brainly.com/question/2137260.

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