What work is done by a forceF =(3.00×N) i^ +(6.00 N) j^ , withxin meters, that moves a particle from a positionr 1 =(2.00 m) i^ +(2.00 m) j^ to a positionr f =−(7.00 m) i^ −(9.00 m) j^ ? Number Units
Question
What work is done by a forceF =(3.00×N) i^ +(6.00 N) j^ , withxin meters, that moves a particle from a positionr 1 =(2.00 m) i^ +(2.00 m) j^ to a positionr f =−(7.00 m) i^ −(9.00 m) j^ ? Number Units
Solution
The work done by a force is given by the dot product of the force vector and the displacement vector. The displacement vector is the final position vector minus the initial position vector.
First, let's calculate the displacement vector:
r_f = -(7.00 m) i^ - (9.00 m) j^ r_i = (2.00 m) i^ + (2.00 m) j^
Displacement (d) = r_f - r_i = [-(7.00 m) - (2.00 m)] i^ + [-(9.00 m) - (2.00 m)] j^ = -(9.00 m) i^ - (11.00 m) j^
Next, we calculate the dot product of the force vector and the displacement vector:
F = (3.00 N) i^ + (6.00 N) j^ d = -(9.00 m) i^ - (11.00 m) j^
F . d = (3.00 N * -9.00 m) + (6.00 N * -11.00 m) = -27.00 J - 66.00 J = -93.00 J
So, the work done by the force is -93.00 Joules.
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