A mysterious object is accelerated in a straight line across the table by a myst
ID: 1442473 • Letter: A
Question
A mysterious object is accelerated in a straight line across the table by a mysterious force. The force is not constant, but varies depending on the position of the object. The graph and equations below give the magnitude of the force F on the object as a function of the position x of the object. Find the work Wo done by the force on the object as it travels from x=2.00 mm to x=9.00 mm.
Fr) N] 20 N 100 x+ 60 N, 0 mm2 mm 2 mm 110 N 40 N 80 60 N x, 2 mmSxS6 mm 4 mm 60 40 x- 40 N, 6 mm5 xs 10 mm 4 mm Number 4 x [mm] tn m 0Explanation / Answer
The work is equal to the area between the graph and the x axis. To determine the area from x = 2mm to x = 9mm, I drew two vertical lines. The lines go from the x = 2 to the graph and x = 9 to the graph. Then I drew a vertical line from x = 6 to the graph. Now I see two trapezoids.
From x = 2 to x = 6, the force decreased from 80 N to 20 N. Let’s use the following equation to determine the work.
Work = ½ * (Fi + Ff) * d
d = 4 mm = 0.004 meter
Work = ½ * (80 + 20) * 0.004 = 0.2 N *m
From 6 mm to 10 mm, the force increased from 20 N to 60 N. Let’s use the following equation to determine the slope of this line.
Slope = F/ x = 40/4= 10
Let’s use the following equation to determine the force at 9 mm.
F = 20 + 10 * x, x = 3
F = 20 + 3 * 10 = 50 N
Work = ½ * (20 + 50) * 0.003 = 0.105 N *m
Total work = 0.2 + 0.105 = 0.305 N * m
Hope it is correct.
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