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A machine carries a 4.0 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 = (7.20 m) + (13.0 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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krlosdark12

Answer:

a) [tex]W=105J[/tex]

b) [tex]P=9.5W[/tex]

Explanation:

The work is defined as:

[tex]W=F*d[/tex]

when the force is on the same direction of the movement.

We need to calculate the displacement in each direction, so:

[tex]\Delta x=x_f-x_i\\\Delta x=7.20m-0.50m\\\Delta x=6.70m\\\\\Delta y=y_f-y_i\\\Delta y=13.0m-0.75m\\\Delta y=12.25m\\\\\Delta z=z_f-z_i\\\Delta z=7.30m-0.20m\\\Delta z=7.10m\\\\[/tex]

now having the displacement, we can calculate the work done by the machine:

[tex]W_x=F_x*\Delta x\\W_x=2.00N*6.70m\\W_x=13.4J\\\\W_y=F_y*\Delta y\\W_y=4.00N*12.25m\\W_y=49.0J\\\\W_z=F_z*\Delta z\\W_z=6.00N*7.10m\\W_z=42.6J[/tex]

The total work is the sum of each of them:

[tex]W=W_x+W_y+W_z\\W=13.4J+49J+42.6J\\W=105J[/tex]

and the power is given by:

[tex]P=\frac{W}{t}\\\\P=\frac{105J}{11s}\\\\P=9.5W[/tex]

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