A mysterious object is accelerated in a straight line across the table by a myst

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 W_o done by the force on the object as it travels from x = 2.00 mm to x = 10.00 mm. F(x) {20 N/2 mm x + 60 N, 0 mm lessthanorequalto x lessthanorequalto 2 mm 110 N - 60 N/4 mm x, 2 mm lessthanorequalto x lessthanorequalto 6 mm 40 N/4 mm x - 40 N, 6 mm lessthanorequalto x lessthanorequalto 10 mm W_o

Explanation / Answer

The work is equal to the area between the graph and the x axis. To determine the area from x = 2 mm to x = 10 mm, I drew two vertical lines. The lines go from the x = 2 to the graph and x = 10 to the graph. Then I drew a vertical line from x = 6 to the graph. Now I see two trapezoids.



From 2 mm to 6 mm, the force decreased from 80 N to 20 N. Let’s use the following equation to determine the slope of this line.

Slope = F ÷ x = -60 ÷ 4 = -15

Let’s use the following equation to determine the force at 4 mm.
F = 80 – 15* x, x = 2
F = 80 – 15 * 2 = 50 N
From x = 4 to x = 6, the force decreased from 50 N to 20 N. Let’s use the following equation to determine the work.

Work = ½ * (Fi + Ff) * d
d = 2 mm = 0.002 meter
Work = ½ * (50 + 20) * 0.002 = 0.07 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 10 mm.
F = 20 + 4 * x, x = 5
F = 20 + 3 * 10 = 50 N

Work = ½ * (20 + 50) * 0.005 = 0.175 N *m
Total work = 0.07 + 0.175 = 0.245 J

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