Turning points: 0.5m and 3.5m.
Speed at 2m: 4.9m/s.
Max speed: 6.9m/s at 2m.
Motion description: Starts, slows to turn at 3.5m, then speeds up to max at 2m, slows and turns back at 0.5m.
Lower energy motion: Oscillates between new turning points closer to origin and 2m.
Here's the breakdown of your questions:
a) Turning Points:
The turning points occur where the potential energy curve intersects the particle's total mechanical energy (12 J) line. Looking at the graph, these points are approximately:
Left turning point: x ≈ 0.5 m
Right turning point: x ≈ 3.5 m
These are the positions where the particle's kinetic energy drops to zero, causing it to change direction.
b) Speed at x = 2.0 m:
To find the particle's speed at x = 2.0 m, we need to calculate its kinetic energy at that point. Since the total mechanical energy is 12 J, and the potential energy at x = 2.0 m is approximately 6 J (from the graph), the kinetic energy is:
Kinetic energy = Total energy - Potential energy = 12 J - 6 J = 6 J
Now, using the formula for kinetic energy:
Kinetic energy = 1/2 * mass * velocity^2
Rearranging for velocity:
velocity = sqrt(2 * kinetic energy / mass) = sqrt(2 * 6 J / 0.5 kg) ≈ 4.9 m/s
Therefore, the particle's speed at x = 2.0 m is approximately 4.9 m/s.
c) Maximum Speed and Positions:
The particle's maximum speed occurs where its kinetic energy is at its maximum. This happens when the potential energy is at its minimum, which is at x = 2.0 m. Here, the potential energy is 0 J, so all the total energy (12 J) is kinetic energy. Plugging this into the kinetic energy formula:
velocity = sqrt(2 * 12 J / 0.5 kg) ≈ 6.9 m/s
Therefore, the particle's maximum speed is approximately 6.9 m/s and occurs at x = 2.0 m.
d) Particle Motion Description:
Starting from the left turning point (x ≈ 0.5 m), the particle has non-zero kinetic energy and moves to the right. As it approaches x = 2.0 m, its potential energy increases gradually, reducing its kinetic energy. At x = 2.0 m, the particle has its maximum speed (kinetic energy is at its maximum). Beyond this point, the potential energy increases more rapidly, further reducing the particle's kinetic energy until it reaches zero at the right turning point (x ≈ 3.5 m). Here, the particle comes to a stop and changes direction.
e) Motion with Lowered Energy:
- If the particle's energy is lowered to 4.0 J, its trajectory is confined to a narrower region within the potential energy curve. This means:
- The left turning point moves closer to the origin (around x ≈ 0.2 m) where the 4 J line intersects the potential energy curve.
- The right turning point moves closer to x = 2.0 m (but stays to the left of it) where the 4 J line intersects the potential energy curve again.
- The particle will oscillate between these new turning points, never reaching x = 2.0 m or beyond.
- Its maximum speed will be lower than before, as the total energy available for kinetic energy is smaller.
The probable question is in the image attached.