Figure 10.21 shows a particle’s potential energy as a function of its position. When is the force on the particle nonzero and only pointing in the negative x-direction? (Choose all that apply).

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To analyze the force from the potential energy plot, recall that the force is F = -dU/dx. In one dimension, this means: - If dU/dx > 0 (the potential energy increases with x), then F is negative (points in the -x direction). - If dU/dx < 0 (the potential energy decreases with x), then F is positive (points in the +x direction). - If dU/dx = 0, the force is zero. With this in mind, we evaluate each interval based on whether the potential energy is rising or falling there. Option 1: Between positions x2 and x3 If the plot shows U increasing as x goes from x2 to x3, then dU/dx > 0 and F < 0, i.e., the force is nonzero and directed in the negative x-direction. If, however, U were decreasing in this interval, the force would be positive. Since the provided answer indicates this interval as correct, we assume the segment from x2 to x3 has an increasing potential energy, yielding a negative force. Option 2: Between positions x3 and x4 Here, we would examine the slope of U from x3 to x4. If the potential energy decreases over this interval (dU/dx < 0), then F = -dU/dx would be positive, meaning the force points in the +x direction, not the -x direction. If U is flat (dU/dx = 0), the force would be zero. Since this option is not listed among the correct answers in the provided data, the slope in this interval is interpreted as not giving a negative, nonzero force. Option 3: Before position x2 In the region to the left of x2, assess whether U is increasing with x. If dU/dx > 0 here, then F < 0 and the force would be nonzero and negative. If instead U decreases with x in this region, the force would be positive. Since the stated correct options do not include this interval, it is inferred that either dU/dx ≤ 0 or the slope is zero in this region. Option 4: After position x4 If the potential energy increases as x increases beyond x4 (dU/dx > 0), then F is negative and nonzero, pointing in -x. The provided correct answer includes this interval, so we conclude that after x4 the slope is positive, giving a nonzero force in the negative x-direction. Summary of reasoning across options: - The key criterion is dU/dx > 0 to produce a nonzero force pointing in the -x direction. - Options are evaluated by the sign of the slope of U in each interval, and whether the slope is zero or negative would not satisfy the condition for a negative, nonzero force. - Based on the given answer set, the intervals that meet the criterion are between x2 and x3 (positive slope of U) and after x4 (also a positive slope of U). The intervals before x2 and between x3 and x4 do not produce a negative nonzero force according to the provided solution, either due to a negative slope (giving a positive force) or a zero slope (no force).

PHYS 2305 Spring 25 Lin Quiz 07 - Chapter 9 (9.4-9.6) and Chapter 10 (10.1-10.8)

Figure 10.21 shows a particle’s potential energy as a function of its position. When is the force on the particle nonzero and only pointing in the negative x-direction? (Choose all that apply).

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