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A force vector F = (4.54 N)i + (6.01 N)j acts on a particle as it moves along the x-axis from x = 3.27 m to x = 7.28 m.What is the work done by the force F as the particle moves along the x-axis in this interval?

A) 18.51 J
B) 24.99 J
C) 33.55 J
D) 29.07 J

1 Answer

4 votes

The work done by the force F as the particle moves along the x-axis from x = 3.27 m to x = 7.28 m is found by multiplying the x-component of the force by the displacement, resulting in 18.21 J.

To calculate the work done by a force vector F as a particle moves along the x-axis, we must apply the concept of work in physics, which is the dot product of the force vector and the displacement vector. Since the force vector is F = (4.54 N)i + (6.01 N)j, and considering the displacement is purely along the x-axis, only the x-component of the force does work.

From the initial position x = 3.27 m to the final position x = 7.28 m, the displacement Δx is 7.28 m - 3.27 m = 4.01 m. The work done by the force F is therefore W = Fx * Δx = 4.54 N * 4.01 m = 18.21 J.

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