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Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 m is
A
13 J
B
18 J
C
21 J
D
26 J
Detailed Solution
Work done = area under the force–distance graph.
From the graph, the force is 2 N from d = 3 m to d = 7 m and then falls uniformly to zero at d = 12 m.
Area of the rectangle (3 m to 7 m) = $2\times(7 - 3) = 8$ J
Area of the triangle (7 m to 12 m) = $\frac{1}{2}\times(12 - 7)\times2 = 5$ J
Total work = 8 + 5 = 13 J
From the graph, the force is 2 N from d = 3 m to d = 7 m and then falls uniformly to zero at d = 12 m.
Area of the rectangle (3 m to 7 m) = $2\times(7 - 3) = 8$ J
Area of the triangle (7 m to 12 m) = $\frac{1}{2}\times(12 - 7)\times2 = 5$ J
Total work = 8 + 5 = 13 J