A machine carries a 3.0 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 = (7.30 m) + (11.0 m) + (7.20 m) at t = 14 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
(a)96.6J
(b)6.9W
To find the work done on the package by the machine's force, we can use the equation:
Work = Force * Displacement * cos(theta)
Where theta is the angle between the force and displacement vectors.
For part (a), we can calculate the work done as follows:
1. Calculate the displacement vector:
Displacement, d = df - di = (7.30 - 0.50) m i + (11.0 - 0.75) m j + (7.20 - 0.20) m k
d = 6.8 m i + 10.25 m j + 7.0 m k
2. Calculate the angle between the force and displacement vectors:
The dot product of the force and displacement vectors gives us the product of their magnitudes and the cosine of the angle between them:
Force dot Displacement = |Force| * |Displacement| * cos(theta)
F dot d = (2.00 N * 6.8 m) + (4.00 N * 10.25 m) + (6.00 N * 7.0 m)
F dot d = 13.6 N*m + 41 N*m + 42 N*m
F dot d = 96.6 N*m
3. Finally, calculate the work done:
Work = F dot d = 96.6 N*m
Therefore, the work done on the package by the machine's force is 96.6 joules (J).
Now, let's move on to part (b) to find the power of the machine's force on the package.
Power is the rate at which work is done, and it can be calculated using the formula:
Power = Work / Time
In this case, the time is given as 14 s, and the work done is 96.6 J. Therefore:
Power = 96.6 J / 14 s
Power = 6.9 watts (W)
So, the power of the machine's force on the package is 6.9 watts.