shows a plot of potential energy U versus position x of a 0.310 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA = 9 J, UC = 20 J and UD = 24 J. The particle is released at the point where U forms a “potential hill” of “height” UB = 12 J, with kinetic energy 4.50 J. What is the speed of the particle at (a)x = 3.5 m and (b)x = 6.5 m? What is the position of the turning point on (c) the right side and (d) the left side?

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To find the speed of the particle at two different positions, as well as the positions of the turning points, we can use the conservation of mechanical energy.

Given information:
- Mass of the particle (m): 0.310 kg
- Potential energy at point A (UA): 9 J
- Potential energy at point C (UC): 20 J
- Potential energy at point D (UD): 24 J
- Potential energy at point B (UB): 12 J
- Kinetic energy at the point of release (KE): 4.50 J
- Position at point x1: 3.5 m
- Position at point x2: 6.5 m

Step 1: Calculate the total mechanical energy at the point of release (E1)
E1 = KE + UB
E1 = 4.50 J + 12 J
E1 = 16.50 J

Step 2: Calculate the total mechanical energy at point x1 (E2)
E2 = U(x1) + KE
Here, U(x1) represents the potential energy at point x1 on the graph.

Since the graph shows a "potential hill" at point B, the potential energy decreases as the particle moves to the right. Therefore, U(x1) = UB + (UA - UB) = 12 J + (9 J - 12 J) = 9 J.

E2 = 9 J + 4.50 J
E2 = 13.50 J

Step 3: Calculate the total mechanical energy at point x2 (E3)
E3 = U(x2) + KE
Here, U(x2) represents the potential energy at point x2 on the graph.

Since the graph shows a "potential hill" of height UD at point D, the potential energy decreases as the particle moves to the left. Therefore, U(x2) = UD + (UC - UD) = 24 J + (20 J - 24 J) = 20 J.

E3 = 20 J + 4.50 J
E3 = 24.50 J

Step 4: Calculate the speed of the particle at point x1 (v1)
Using the conservation of mechanical energy (E1 = E2) and the equation for kinetic energy (KE = 0.5mv^2), we can solve for v1.

E1 = E2
0.5mv1^2 + UB = U(x1) + KE
0.5 * 0.310 kg * v1^2 + 12 J = 9 J + 4.50 J

Simplifying the equation:
0.155 v1^2 + 12 J = 13.50 J
0.155 v1^2 = 1.50 J
v1^2 = 1.50 J / 0.155
v1^2 = 9.6774
v1 ≈ √9.6774
v1 ≈ 3.11 m/s

Step 5: Calculate the speed of the particle at point x2 (v2)
Using the conservation of mechanical energy (E1 = E3) and the equation for kinetic energy (KE = 0.5mv^2), we can solve for v2.

E1 = E3
0.5mv2^2 + UB = U(x2) + KE
0.5 * 0.310 kg * v2^2 + 12 J = 20 J + 4.50 J

Simplifying the equation:
0.155 v2^2 + 12 J = 24.50 J
0.155 v2^2 = 12.50 J
v2^2 = 12.50 J / 0.155
v2^2 = 80.6452
v2 ≈ √80.6452
v2 ≈ 8.98 m/s

Step 6: Calculate the position of the turning points
(a) Right side turning point:
The right side turning point is the position where the particle stops momentarily before changing direction. This occurs when the kinetic energy becomes zero.

At the right side turning point, KE = 0.
E = U(xright) + KE
E = UC + KE
UC + 0 = UC + 0.5mvright^2
vright^2 = 0
Therefore, the speed of the particle at the right side turning point is zero.

(b) Left side turning point:
The left side turning point is the position where the particle stops momentarily before changing direction. This occurs when the kinetic energy becomes zero.

At the left side turning point, KE = 0.
E = U(xleft) + KE
E = UA + KE
UA + 0 = UA + 0.5mvleft^2
vleft^2 = 0
Therefore, the speed of the particle at the left side turning point is zero.

In summary:
(a) The speed of the particle at x = 3.5 m is approximately 3.11 m/s.
(b) The speed of the particle at x = 6.5 m is approximately 8.98 m/s.
(c) The right side turning point has a speed of 0 m/s.
(d) The left side turning point has a speed of 0 m/s.

To solve this problem, we will use the conservation of mechanical energy principle. At any point in the system, the sum of the kinetic energy (KE) and potential energy (PE) is constant.

Let's analyze the given information step-by-step:

Step 1: Determine the total mechanical energy at the starting point.
At point B, where the particle is released, its kinetic energy is given as KE = 4.50 J. This means the potential energy at point B is UB = 12 J (the remaining energy component), as the sum of KE and PE is constant.
So, at point B, the total mechanical energy is the sum of kinetic and potential energy: E_total = KE + PE = 4.50 J + 12 J = 16.50 J.

Step 2: Find the speed of the particle at x = 3.5 m.
At any point, the mechanical energy, E_total, remains constant. Using the given values of potential energy (UA, UC, UD) and mechanical energy at point B (E_total), we can determine the potential energy at x = 3.5 m (U3.5) as follows:
U3.5 = E_total - UA = 16.50 J - 9 J = 7.50 J

Since we have the potential energy at x = 3.5 m, we can find the kinetic energy at this point using the conservation of mechanical energy principle:
KE3.5 = E_total - U3.5 = 16.50 J - 7.50 J = 9.00 J

Next, we can calculate the speed (v3.5) at x = 3.5 m by using the formula for kinetic energy:
KE = (1/2)mv^2
v3.5 = sqrt((2*KE3.5)/m) = sqrt((2*9.00 J)/(0.310 kg)) ≈ 8.04 m/s

Therefore, the speed of the particle at x = 3.5 m is approximately 8.04 m/s.

Step 3: Find the speed of the particle at x = 6.5 m.
We will use the same logic as in Step 2 to find the speed (v6.5) at x = 6.5 m.
First, calculate the potential energy (U6.5) at x = 6.5 m:
U6.5 = E_total - UD = 16.50 J - 24 J = -7.50 J

Then, find the kinetic energy (KE6.5) at x = 6.5 m:
KE6.5 = E_total - U6.5 = 16.50 J - (-7.50 J) = 24.00 J

Finally, calculate the speed (v6.5) at x = 6.5 m:
v6.5 = sqrt((2*KE6.5)/m) = sqrt((2*24.00 J)/(0.310 kg)) ≈ 9.71 m/s

Therefore, the speed of the particle at x = 6.5 m is approximately 9.71 m/s.

Step 4: Determine the positions of the turning points (left and right side).
At the turning points, the velocity of the particle becomes zero, indicating a change in direction. This occurs where the kinetic energy is equal to zero.
Using the conservation of mechanical energy, we know that at these points, the total mechanical energy (E_total) is equal to the potential energy (U).
Therefore, the position of the turning points can be determined by finding the potential energy (U) equal to E_total.

(a) Right side turning point:
Given that the total mechanical energy is E_total = 16.50 J, we need to find the position (x) where U = E_total.
UB = 12 J and UC = 20 J, so at the right side turning point, we can use the equation:
UD - U = E_total
24 J - U = 16.50 J
U = 24 J - 16.50 J = 7.50 J

From the potential energy graph, we see that U = 7.50 J corresponds to a position of x = 2.75 m (approximately). Therefore, the position of the right side turning point is approximately x = 2.75 m.

(b) Left side turning point:
The potential energy (U) at the left side turning point is also equal to E_total, which is 16.50 J.
From the given information, we can see that UA = 9 J, so at the left side turning point, we can use the equation:
UA - U = E_total
9 J - U = 16.50 J
U = 9 J - 16.50 J = -7.50 J

Again, from the potential energy graph, we deduce that U = -7.50 J corresponds to a position of x = 9.00 m (approximately). Therefore, the position of the left side turning point is approximately x = 9.00 m.

In summary:
(a) The speed of the particle at x = 3.5 m is approximately 8.04 m/s.
(b) The speed of the particle at x = 6.5 m is approximately 9.71 m/s.
(c) The position of the right side turning point is approximately x = 2.75 m.
(d) The position of the left side turning point is approximately x = 9.00 m.