A position dependent force, F= 7-2x 3x^2 N acts on a small body of mass 2kg and displaces it from x = 0 to x = 5m. The work done in joules is?
Answer: 135
Explanation
To find the work done, we need to integrate the force with respect to the displacement.
Given:
F(x) = 7 - 2x + 3x^2 N
x_initial = 0 m
x_final = 5 m
Work done (W) is the integral of force (F) with respect to displacement (x):
W = ∫[x_initial, x_final] F(x) dx
= ∫[0, 5] (7 - 2x + 3x^2) dx
Evaluating the integral:
W = [7x - x^2 + x^3] from 0 to 5
= (7(5) - 5^2 + 5^3) - (7(0) - 0^2 + 0^3)
= (35 - 25 + 125) - 0
= 135 J
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