A 1.30 kg block slides with a speed of 0.895 m/s on a frictionless horizontal su
ID: 1626989 • Letter: A
Question
A 1.30 kg block slides with a speed of 0.895 m/s on a frictionless horizontal surface until it encounters a spring with a force constant of 605 N/m. The block comes to rest after compressing the spring 4.15 cm. Find the spring potential energy, U, the kinetic energy of the block, K, and the total mechanical energy of the system, E, for compressions of 3.00 cm. Enter your answers numerically separated by commas. Find the spring potential energy, U, the kinetic energy of the block, K, and the total mechanical energy of the system, E, for compressions of 4.00 cm. Enter your answers numerically separated by commas.Explanation / Answer
Assuming elastic collision between the block and the spring,
Kinetic energy of the block gets stored in the spring as potential energy at maxm. compression
=> (1/2) * (1.30) * (0.895)^2 = (1/2) k * (0.0415)^2, where k is spring constant in N/m
=> k = 604.63 N/m
At 3 cm compression:
PE of the spring = (1/2) kx^2 = (1/2) * 604.63 * (0.03)^2 J = 0.27208 J,
Kinetic energy of the block = (1/2) * (1.30) * (0.895)^2 J = 0.52067 J
Mechanical energy of the system = 0.79275 J
At 4.00 cm compression:
PE of the spring = (1/2) kx^2 = (1/2) * 604.63 * (0.04)^2 J = 0.48370 J
Mechanical energy of the system = 0.79275 J
Kinetic energy of the block = 0.79275 - 0.48370 J = 0.30905 J.
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