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Ou attach a 1.30 kg block to a horizontal spring that is fixed at one end. You pull the block until the spring is stretched by 0.400 m and release it from rest. Assume the block slides on a horizontal surface with negligible friction. The block reaches speed of zero again 0.500 s after release (for the first time after release).

a. What is the maximum speed of the block

User Samil
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Final answer:

The maximum speed of the block is found by setting the potential energy of the stretched spring equal to the kinetic energy of the block and solving for velocity. Given the motion is frictionless and harmonic, we could also infer this from the period and frequency.

Step-by-step explanation:

To determine the maximum speed of a block attached to a spring and released from rest, we utilize the principles of conservation of energy in a frictionless system. When the block is released, the potential energy stored in the spring is transferred to kinetic energy of the block. The spring's potential energy at maximum stretch (or compression) is given by PE = 1/2 kx2, where k is the spring constant and x is the displacement from equilibrium.

For the student's scenario, the potential energy stored in the spring at maximum stretch is its maximum kinetic energy due to the frictionless surface. The formula for kinetic energy is KE = 1/2 mv2, where m is the mass of the block and v is its velocity. To find the maximum speed, we set the kinetic energy equal to the potential energy at maximum stretch.

Solving for v, we obtain v = sqrt((k/m) * x2). Given the time to reach speed zero again, we can also deduce that it underwent simple harmonic motion and reached maximum speed at the equilibrium position, which is half the time of a full oscillation. Since the block takes 0.500 s to return to rest, the period T of the motion is 1.000 s, and the frequency f is 1/T, which can be related to the spring constant and mass by f = 1/(2π) * sqrt(k/m).

User Yiou
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