A 2.4-kg mass starts from rest at point A and moves along the x axis subject to

Question

A 2.4-kg mass starts from rest at point A and moves along the x axis subject to the potential energy shown in the figure below.

(a) Determine the speed of the mass at points B, C, D.

Point B: ___ m/s

Point C: ___ m/s

Point D: ___ m/s

(b) Determine the turning points for the mass. (Select all that apply.)

[ ] Point A

[ ] Point B

[ ] Point C

[ ] Point D

[ ] Point E

A 2.4-kg mass starts from rest at point A and moves along the x axis subject to the potential energy shown in the figure below. (a) Determine the speed of the mass at points B, C, D. Point B: ___ m/s Point C: ___ m/s Point D: ___ m/s (b) Determine the turning points for the mass. (Select all that apply.) [ ] Point A [ ] Point B [ ] Point C [ ] Point D [ ] Point E

Explanation / Answer

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.

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