The figure shows a plot of potential energy U versus position x of a 0.260 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA = 9.00 J, Uc=20.0 J and Up = 24.0 J. The particle is released at the point where U forms a "potential hill" of "height" Ug = 12.0 J, with kinetic energy 7.50 J. What is the speed of the particle at (a)x= 3.50 m and (b)x= 6.50 m? What is the position of the turning point on (c) the right side and (d) the left side? (a) Number i (b) Number i (c) Number i W 4 5 6 x (m) U(J) Uch UB U₁ 0 lov 2 3 Units Units Units 8 <

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The figure shows a plot of potential energy \( U \) versus position \( x \) of a 0.260 kg particle that can travel only along an \( x \) axis under the influence of a conservative force. The graph has these values: \( U_A = 9.00 \, \text{J}, \, U_C = 20.0 \, \text{J} \) and \( U_D = 24.0 \, \text{J} \). The particle is released at the point where \( U \) forms a “potential hill” of “height” \( U_B = 12.0 \, \text{J} \), with kinetic energy 7.50 J. What is the speed of the particle at (a) \( x = 3.50 \, \text{m} \) and (b) \( x = 6.50 \, \text{m} \)? What is the position of the turning point on (c) the right side and (d) the left side?

**Graph Description:**
The graph displays potential energy \( U(x) \) (in joules) on the y-axis against position \( x \) (in meters) on the x-axis. 
- At \( x = 0 \) to \( x = 2 \, \text{m} \), the potential energy increases linearly from \( U_A = 9.00 \, \text{J} \) to \( U_C = 20.0 \, \text{J} \).
- From \( x = 2 \, \text{m} \) to \( x = 4 \, \text{m} \), it decreases linearly back to \( U_B = 12.0 \, \text{J} \).
- Between \( x = 4 \, \text{m} \) and \( x = 6 \, \text{m} \), the energy remains constant at \( U_B = 12.0 \, \text{J} \).
- Then it increases to \( U = 24.0 \, \text{J} \) at \( x = 7 \, \text{m} \) and stays constant up to \( x = 8 \, \text{m} \).
- Beyond \( x = 8 \, \text
Transcribed Image Text:The figure shows a plot of potential energy \( U \) versus position \( x \) of a 0.260 kg particle that can travel only along an \( x \) axis under the influence of a conservative force. The graph has these values: \( U_A = 9.00 \, \text{J}, \, U_C = 20.0 \, \text{J} \) and \( U_D = 24.0 \, \text{J} \). The particle is released at the point where \( U \) forms a “potential hill” of “height” \( U_B = 12.0 \, \text{J} \), with kinetic energy 7.50 J. What is the speed of the particle at (a) \( x = 3.50 \, \text{m} \) and (b) \( x = 6.50 \, \text{m} \)? What is the position of the turning point on (c) the right side and (d) the left side? **Graph Description:** The graph displays potential energy \( U(x) \) (in joules) on the y-axis against position \( x \) (in meters) on the x-axis. - At \( x = 0 \) to \( x = 2 \, \text{m} \), the potential energy increases linearly from \( U_A = 9.00 \, \text{J} \) to \( U_C = 20.0 \, \text{J} \). - From \( x = 2 \, \text{m} \) to \( x = 4 \, \text{m} \), it decreases linearly back to \( U_B = 12.0 \, \text{J} \). - Between \( x = 4 \, \text{m} \) and \( x = 6 \, \text{m} \), the energy remains constant at \( U_B = 12.0 \, \text{J} \). - Then it increases to \( U = 24.0 \, \text{J} \) at \( x = 7 \, \text{m} \) and stays constant up to \( x = 8 \, \text{m} \). - Beyond \( x = 8 \, \text
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