how much work is done by each force acting on vacuum cleaner as it is pulled horizontally 3 m at a constant speed by a force of 50 N at an angle of 30degrees above the horizontal?
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Dang I needed this answer but every one slow
To determine how much work is done by each force acting on the vacuum cleaner, we need to consider the definition of work, which is given by the formula:
Work = Force × Distance × cos(θ)
Where:
- Force is the magnitude of the force applied (in Newtons),
- Distance is the displacement of the object (in meters), and
- θ is the angle between the force vector and the direction of displacement (in degrees).
In this case, there are two forces acting on the vacuum cleaner:
1. The force applied to pull the vacuum cleaner horizontally, which has a magnitude of 50 N and an angle of 0 degrees with the horizontal direction.
2. The gravitational force acting vertically, which can be represented by the weight of the vacuum cleaner. Since the vacuum cleaner is pulled horizontally, the vertical component of the weight does not contribute to the work done.
Now let's calculate the work done by each force:
1. The work done by the force applied horizontally:
- Force = 50 N
- Distance = 3 m
- θ = 0 degrees (since it's pulled horizontally)
- Work1 = 50 N × 3 m × cos(0°)
Since cos(0°) equals 1, the work done by this force is:
Work1 = 50 N × 3 m × 1 = 150 Joules (J)
2. The work done by the gravitational force (weight):
- The vertical component of the weight does not contribute to the work done, so we only consider the horizontal component.
- Since the vacuum cleaner is moving horizontally at a constant speed (assuming no vertical movement), the vertical component of the weight does not change the displacement.
Thus, the work done by the gravitational force is zero.
Therefore, the work done by the force applied to pull the vacuum cleaner horizontally is 150 Joules, while the work done by the gravitational force is zero.