A machine carries a 4.5 kg package from an initial position of di = (0.50 m) + (0.75 m) + (0.20 m) at t = 0 to a final position of df = (7.60 m) + (12.0 m) + (7.30 m) at t = 11 s. The constant force applied by the machine on the package is F = (2.00 N) + (4.00 N) + (6.00 N) .
(a) For that displacement, find the work done on the package by the machine's force.
J
(b) Also, find the power of the machine's force on the package.
W
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(a) To find the work done on the package by the machine's force, we can use the formula:
Work = Force * Displacement * cos(theta)
where theta is the angle between the force and displacement vectors.
First, let's calculate the displacement vector:
displacement (d) = df - di
=(7.60 m) + (12.0 m) + (7.30 m) - (0.50 m) - (0.75 m) - (0.20 m)
=(7.60 m - 0.50 m) + (12.0 m - 0.75 m) + (7.30 m - 0.20 m)
=(7.10 m) + (11.25 m) + (7.10 m)
=(25.45 m) + (11.25 m)
=(36.70 m)
Next, calculate the dot product of the force and displacement vectors to find the magnitude of the work:
Work = Force * Displacement * cos(theta)
= (2.00 N) + (4.00 N) + (6.00 N) * (36.70 m) * cos(theta)
Since the force and displacement vectors are given in component form, we can find the dot product by multiplying the corresponding components and summing them:
Work = (2.00 N * 36.70 m) + (4.00 N * 36.70 m) + (6.00 N * 36.70 m)
= (73.40 N*m) + (146.80 N*m) + (220.20 N*m)
= 440.40 N*m
Therefore, the work done on the package by the machine's force is 440.40 N*m.
(b) The power of the machine's force on the package can be calculated using the formula:
Power = Work / time
Given that the time (t) is 11 s and the work (W) is 440.40 N*m, we can substitute these values into the formula:
Power = 440.40 N*m / 11 s
= 40 N*m/s
Therefore, the power of the machine's force on the package is 40 W.