A force of 6 N is exerted on an object, resulting in its displacement of 2 m in the direction of the force. If the force is constant during the displacement, what is the work done in this case?
12 J
The work done by a constant force is calculated using the formula:
\(W = F \cdot d \cdot \cos(\theta)\)
where:
- \(W\) is the work done (in Joules)
- \(F\) is the magnitude of the force (in Newtons)
- \(d\) is the magnitude of the displacement (in meters)
- \(\theta\) is the angle between the force and the displacement vectors
In this problem, we are given:
- \(F = 6 \, \text{N}\)
- \(d = 2 \, \text{m}\)
- \(\theta = 0^\circ\) (since the displacement is in the direction of the force)
Substituting these values into the formula, we get:
\(W = 6 \, \text{N} \cdot 2 \, \text{m} \cdot \cos(0^\circ)\)
\(W = 12 \, \text{J}\)
Therefore, the work done is 12 Joules.
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