1) An object moves along the x axis, subject to the potential energy shown in th

Question

1)An object moves along the

x axis, subject to the potential energy shown in the figure. (Figure 1) The object has a mass of 2.2kg and starts at rest at point A.

http://session.masteringphysics.com/problemAsset/1123696/2/Walker4e.ch08.Pr055.jpg

What is the object's speed at point B?

What is the object's speed at point C?

What is the object's speed at point D?

What are the turning points for this object?

2) Calculate the work done by gravity as a 3.6

kg object is moved from point A to point B in the figure(Figure 1) along paths 1, 2, and 3.

http://i214.photobucket.com/albums/cc30/bkoz319/08-15.gif

Please help me with these two problems.

An object moves along the x axis, subject to the potential energy shown in the figure. (Figure 1) The object has a mass of 2.2kg and starts at rest at point A. What is the object's speed at point B? What is the object's speed at point C? What is the object's speed at point D? What are the turning points for this object? Calculate the work done by gravity as a 3.6kg object is moved from point A to point B in the figure(Figure 1) along paths 1, 2, and 3.

Explanation / Answer

An object moves along the x axis, subject to the potential energy shown in the figure. The object has a mass of 2.7kg and starts at rest at point A. What is the object's speed at point B, C, and D? What are the turning points for this object?

This problem shouldn't be too difficult. If it makes you feel better by thinking physically, think of a hill in exactly the shape indicated on the graph. AND then just think of horizontal position as being of interest.


We assume that no non-conservative forces act on the object. Thus, between any pair of points, change in potential energy MUST equal the opposite of change in kinetic energy as per energy conservation.

For any points, let's choose that point P and A. No point P is labeled on the graph, just think of it as a placeholder label.

Thus:
(KE_P - KE_A) = (PE_A - PE_P)

Since at point A it is at rest, KE_A = 0

Solve for KE_P:
KE_P = PE_A - PE_P

Remember how KE corresponds to velocity:
KE_P = 1/2*m*vp^2

Thus:
1/2*m*vp^2 = (PE_A - PE_P)

Solve for vp:
vp = sqrt(2*(PE_A - PE_P)/m)


Now, adapt subscripts B, C, and D:
vb = sqrt(2*(PE_A - PE_B)/m)
vc = sqrt(2*(PE_A - PE_C)/m)
vd = sqrt(2*(PE_A - PE_D)/m)

Data:
PE_A := 10 J; PE_B:=2 J; PE_C := 6 J; PE_D:=5 J; m:=2.7 kg;

Results:
vb = 2.434 m/s
vc = 1.721 m/s
vd = 1.925 m/s

As for "turning points", that is simply anywhere where the object will come to a halt, have zero kinetic energy (thus the same potential energy as it had at location A), and be able to reach it within the "potential energy well" from which it began.

Thus, the turning points are A and E.

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