Final answer:
The block is moving at approximately 24.8 m/s after being released from the spring, calculated by using the conservation of energy principle, where the potential energy of the compressed spring is converted into the kinetic energy of the block.
Step-by-step explanation:
The student wants to know how fast a 1.3 kg block is moving after it is released from a compressed spring. The spring has a force constant of 2.0×10⁴ N/m and is compressed by 0.20 m. To solve this, we can use the conservation of energy principle where the potential energy stored in the compressed spring is completely converted into the kinetic energy of the block when it is released and the spring returns to its normal length.
The potential energy (PE) stored in the spring when compressed is given by the formula PE = (1/2)kx², where 'k' is the spring constant and 'x' is the compression distance. Plugging in the given values, PE = (1/2)(2.0×10⁴ N/m)(0.20 m)² = 400 J.
The kinetic energy (KE) of the block when it leaves the spring is equal to the potential energy stored in the spring, so KE = PE = 400 J. The formula for kinetic energy is KE = (1/2)mv², where 'm' is the mass of the block and 'v' is its velocity. We can solve for 'v' by rearranging the formula to v = √(2×KE/m). After substituting the known values, v = √(2× 400 J / 1.3 kg) = √(800/1.3) ≈ √615.38.
Therefore, the velocity of the block when it leaves the spring is ≈ 24.8 m/s.