Question

question 1
1 pts

an object with a mass of 2.74kg is bound by a force described by the potential in the graph above, which each tick on the vertical axis represents 1 joule, and each tick on the horizontal axis one meter. the object reaches a maximum velocity of 1.31m/s during its journey back - and - forth along the x - axis. find the spatial separation of the two turnaround points for this object.
provide at least two decimal places

Explanation:

Step1: Calculate the total mechanical energy

The total mechanical energy \(E\) is the sum of kinetic energy \(K\) and potential energy \(U\). At the point of maximum velocity, the potential energy \(U = 0\) (since velocity is maximum when \(F=-\frac{dU}{dx}=0\), and from the graph, the minimum of \(U(x)\) is the equilibrium point).
The kinetic energy formula is \(K=\frac{1}{2}mv^{2}\). Given \(m = 2.74\space kg\) and \(v=1.31\space m/s\), we have \(K=\frac{1}{2}\times2.74\times(1.31)^{2}\).

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So, the total mechanical energy \(E = 2.351057\space J\) (since \(U = 0\) at maximum - velocity point).

Step2: Find the turnaround points

The turnaround points occur when \(K = 0\) (i.e., \(E=U(x)\)).
From the graph, assume the potential - energy function. The potential - energy function has a "V - shape". Let's assume the left - hand side of the potential - energy function is \(U(x)=a(x - x_0)\) (a linear function). Since \(E = 2.351057\space J\), and each vertical tick is \(1\space J\).
Counting the number of horizontal ticks from the center (minimum of \(U(x)\)) to the turnaround point.
Let's assume the potential - energy function \(U(x)\) has a slope. The total mechanical energy \(E\) is used to find the \(x\) - values where \(U(x)=E\).
Since the potential - energy graph is symmetric about the \(y\) - axis (minimum at \(x = 0\)). Let the left - hand side turnaround point be \(x=-d\) and the right - hand side be \(x = d\).
The spatial separation \(D=2d\).
From the energy conservation \(E = 2.35\space J\) (approximate value for calculation from the first step). Counting the horizontal steps (since each horizontal tick is \(1\space m\)).
If we assume the potential - energy function \(U(x)\) has a slope such that for the left - hand side \(U(x)=-kx\) (for \(x\lt0\)) and \(U(x)=kx\) (for \(x\gt0\)). Since \(E = kd\) (at \(x = d\) or \(x=-d\)).
From the first step \(E\approx2.35\space J\). Counting the number of horizontal units from \(x = 0\) to the point where \(U(x)=E\).
The spatial separation \(D\) is the distance between \(x=-d\) and \(x = d\). Since \(E\) corresponds to \(d\approx2.35\space m\) (because each vertical tick is \(1\space J\) and the potential - energy function is linear in the regions away from the minimum). So \(D = 2\times2.35=4.70\space m\)

Answer:

\(4.70\space m\)