Final answer:
The potential energy stored in the block-spring support system when the block is just released is 0.72 J. The speed of the block when it crosses the point where the spring is neither compressed nor stretched is approximately 1.663 m/s. The speed of the block when it has traveled a distance of 20 cm from where it was released is approximately 2.309 m/s.
Step-by-step explanation:
(a) How much potential energy was stored in the block-spring support system when the block was just released?
The potential energy stored in the block-spring support system when the block is just released can be calculated using the formula:
Potential Energy = 0.5 * k * x^2
Where k is the spring constant and x is the displacement of the spring from its equilibrium position. In this case, k = 100 N/m and x = 12 cm = 0.12 m. Plugging in these values, we get:
Potential Energy = 0.5 * 100 N/m * (0.12 m)^2 = 0.72 J
Therefore, the potential energy stored in the block-spring support system when the block is just released is 0.72 J.
(b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched.
When the block crosses the point where the spring is neither compressed nor stretched, all of the potential energy stored in the spring is converted to kinetic energy.
Using the conservation of energy principle, we can equate the potential energy at the initial position to the kinetic energy at the final position:
Potential Energy = Kinetic Energy
0.5 * k * x^2 = 0.5 * m * v^2
Where m is the mass of the block and v is its velocity. Rearranging the equation and solving for v, we get:
v = sqrt((k * x^2) / m)
Plugging in the values, we have:
v = sqrt((100 N/m * (0.12 m)^2) / 0.3 kg) ≈ 1.663 m/s
Therefore, the speed of the block when it crosses the point where the spring is neither compressed nor stretched is approximately 1.663 m/s.
(c) Determine the speed of the block when it has traveled a distance of 20 cm from where it was released.
To determine the speed of the block when it has traveled a distance of 20 cm from where it was released, we need to consider the change in potential and kinetic energy.
Initially, the potential energy is 0.72 J (as calculated above).
When the block has traveled a distance of 20 cm = 0.2 m, the potential energy is:
Potential Energy = 0.5 * k * x^2 = 0.5 * 100 N/m * (0.2 m)^2 = 0.2 J
Using the conservation of energy principle, we can equate the initial potential energy to the sum of the final potential energy and final kinetic energy:
0.72 J = 0.2 J + 0.5 * m * v^2
Simplifying the equation and solving for v, we get:
v = sqrt((2 * (0.72 J - 0.2 J)) / 0.3 kg) ≈ 2.309 m/s
Therefore, the speed of the block when it has traveled a distance of 20 cm from where it was released is approximately 2.309 m/s.