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svet-max [94.6K]
3 years ago
12

A block is attached to a spring, with spring constant k, which is attached to a wall. It is initially moved to the left a distan

ce d (at point A) and then released from rest, where the block undergoes harmonic motion. The floor is frictionless.The points labelled A and C are the turning points for the block, and point B is the equilibrium point.1)Which of these quantities are conserved for the spring and block system?mechanical energyx-direction linear momentumneither energy nor momentum
Physics
1 answer:
Burka [1]3 years ago
7 0

Answer:

Explanation:

When a spring attached with a block is stretched and released , the block starts moving under SHM. The elastic energy stored in it initially at the time of initial stretch is  repeatedly converted into kinetic energy and vice - versa while the body keeps moving under SHM. So we can tell that the mechanical energy of the system remains constant.

In the whole process the velocity of the body keeps changing in terms of both magnitude and direction . It happens because a variable force acts on the body constantly towards the equilibrium point  so its momentum also keeps changing all the time

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Explanation:

Given:

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To find v₀, use a kinematic equation that's independent of time.

v² = v₀² + 2aΔx

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100 = v₀² − 4000

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v₀ = ±64.0

As your teacher said, v₀ can't be +64.0.  So v₀ = -64.0.

Next, to find time, use a kinematic equation that's independent of initial velocity.

Δx = vt − ½ at²

-100 = (10) t − ½ (20) t²

-100 = 10 t − 10 t²

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