Final answer:
The potential energy stored in the block-spring support system when the block was just released is 0.72 J.
The speed of the block when it crosses the point when the spring is neither compressed nor stretched is 2.39 m/s. The speed of the block when it has traveled a distance of 20 cm from where it was released is 3.46 m/s.
Step-by-step explanation:
(a) The potential energy stored in the block-spring support system when the block was 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 compression or extension of the spring. In this case, the spring was compressed by 12 cm, which is equal to 0.12 m, and the spring constant is 100 N/m. Plugging in these values, we get:
Potential energy = 0.5 * 100 * (0.12)^2 = 0.72 J
(b) The speed of the block when it crosses the point when the spring is neither compressed nor stretched can be calculated using the principle of conservation of mechanical energy. At the point where the spring is neither compressed nor stretched, the potential energy is zero. Therefore, the initial potential energy is equal to the final kinetic energy:
0.5 * m * v^2 = 0.72 J
Solving for v, we get:
v = sqrt(2 * (Potential energy / m)) = sqrt(2 * 0.72 / 0.3) = 2.39 m/s
(c) The speed of the block when it has traveled a distance of 20 cm from where it was released can also be calculated using the principle of conservation of mechanical energy. At this point, the potential energy is zero and the block has only kinetic energy:
0.5 * m * v^2 = 0.5 * k * x^2
where x is the distance traveled by the block, k is the spring constant, and v is the speed of the block. Plugging in the values, we get:
0.5 * 0.3 * v^2 = 0.5 * 100 * (0.2)^2
Solving for v, we get:
v = sqrt((k * x^2) / m) = sqrt((100 * (0.2)^2) / 0.3) = 3.46 m/s