Question

An object with a mass of 0.35 kg moves along the x axis under the influence of one force whose potential energy is given by the graph. In the graph, the vertical spacing between adjacent grid lines represents an energy difference of 5.59 J, and the horizontal spacing between adjacent grid lines represents a displacement of a.

a) What is the maximum speed (in m/s) of the object at x = 6a so that the object is confined to the region 4a < x < 8a?

b)What is the maximum speed (in m/s) of the object at x = 6a so that the object is confined to the region 3a < x < 9a?

c) Suppose the object is at x = 6a and moving in the negative x direction with a speed that is 17.2% greater than that calculated in part (b). What will its speed (in m/s) be when it is at x = 0?

Answer 1

a) The maximum speed of the object at x = 6a so that it is confined to the region 4a < x < 8a is the speed at the point where the potential energy curve intersects the horizontal lines representing the energy levels of 16.77 J and 5.59 J. This occurs at x = 5a and x = 7a, respectively. The change in potential energy between these points is:

ΔU = U(5a) - U(7a) = (3)(5.59 J) = 16.77 J

Using the conservation of energy principle, the maximum kinetic energy and speed of the object at x = 6a is:

KE = ΔU = 16.77 J

(1/2)mv^2 = 16.77 J

v = sqrt((2*16.77 J)/(0.35 kg)) = 8.19 m/s

b) The maximum speed of the object at x = 6a so that it is confined to the region 3a < x < 9a is the speed at the point where the potential energy curve intersects the horizontal lines representing the energy levels of 16.77 J and 0 J. This occurs at x = 3.5a and x = 8.5a, respectively. The change in potential energy between these points is:

ΔU = U(3.5a) - U(8.5a) = (6)(5.59 J) = 33.54 J

Using the conservation of energy principle, the maximum kinetic energy and speed of the object at x = 6a is:

KE = ΔU = 33.54 J

(1/2)mv^2 = 33.54 J

v = sqrt((2*33.54 J)/(0.35 kg)) = 12.36 m/s

c) The change in potential energy between x = 6a and x = 0 is:

ΔU = U(6a) - U(0) = (10)(5.59 J) = 55.9 J

Let v' be the speed of the object when it is at x = 0. The conservation of energy principle gives:

(1/2)mv^2 - ΔU = (1/2)mv'^2

(1/2)(0.35 kg)(1.172v)^2 - 55.9 J = (1/2)(0.35 kg)(v')^2

Solving for v', we get:

v' = sqrt[((1/2)(0.35 kg)(1.172v)^2 - 55.9 J)/(0.5*0.35 kg)] = 6.05 m/s

Answer 2

The maximum speed of an object at a specific point can be determined through conservation of energy, equating kinetic and potential energies at various points. Specific calculations require the potential energy versus position graph which is not provided. The principle involves conservation of energy where no energy is lost to external factors.

Step-by-step explanation:

To determine the maximum speed of the object at a particular point with specified conditions, we need to use the principle of conservation of energy. The total mechanical energy (kinetic plus potential) at one point in time will equal the total mechanical energy at any other point, provided no external work is done on the system.

While a specific graph of potential energy versus position is mentioned in the student's question, without the actual graph or data from the graph, we cannot calculate numerical values. Therefore, we address the concepts involved.

• To find the maximum speed of the object at x = 6a that won't allow it to exceed the region 4a < x < 8a, we must calculate the potential energy at the limits of this region and equate it to the sum of the potential energy and kinetic energy at x = 6a.
• For the region 3a < x < 9a, we do a similar calculation considering the potential energy at these new limits.
• If the object at x = 6a is moving in the negative x direction with a speed 17.2% greater than the one obtained in part b, its new kinetic energy can be calculated. When the object reaches x = 0, all of this kinetic energy will be converted into potential energy, assuming no energy is lost.

In cases like these, the use of the formula Kinetic Energy = ½ m * v² and Potential Energy = m * g * h (where m is the mass, v is the velocity, g is the acceleration due to gravity, and h is the height) is typically involved, along with understanding that the force can be derived as the negative gradient of the potential energy curve.