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A force F acting on an object varies with distance $x$ as shown here.
(F = 3 N from $x$ = 0 to $x$ = 3 m, and then decreases linearly to zero at $x$ = 6 m.)
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The force is in N and $x$ in m. The work done by the force in moving the object from $x$ = 0 to $x$ = 6 m is
(F = 3 N from $x$ = 0 to $x$ = 3 m, and then decreases linearly to zero at $x$ = 6 m.)
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The force is in N and $x$ in m. The work done by the force in moving the object from $x$ = 0 to $x$ = 6 m is
A
18.0 J
B
13.5 J
C
4.5 J
D
9.0 J
Detailed Solution
Work done by a variable force: $W = \int F\,dx$ = area under the F-x graph.
From $x$ = 0 to 3 m the force is constant at 3 N: area of the rectangle $= 3 \times 3 = 9$ J
From $x$ = 3 m to 6 m the force falls linearly from 3 N to 0: area of the triangle $= \dfrac{1}{2} \times 3 \times 3 = 4.5$ J
Total work $W = (3 \times 3) + \dfrac{1}{2} \times (3)(3)$
$W = 9 + 4.5 = 13.5$ J
From $x$ = 0 to 3 m the force is constant at 3 N: area of the rectangle $= 3 \times 3 = 9$ J
From $x$ = 3 m to 6 m the force falls linearly from 3 N to 0: area of the triangle $= \dfrac{1}{2} \times 3 \times 3 = 4.5$ J
Total work $W = (3 \times 3) + \dfrac{1}{2} \times (3)(3)$
$W = 9 + 4.5 = 13.5$ J