Question 46

Mathematics Scalar and Vector Products Medium

If \(|F| = 40 N\), \(|D| = 3m\), and \(\theta = 60^\circ\), then the work done by \(F\) acting from \(P\) to \(Q\) is:

(A) \(60\sqrt{3} J\)
(B) 120 J
(C) \(60\sqrt{2} J\)
(D) 60 J
View Dynamic Solution & Explanation
Correct Solution: Option D

Step-by-step Solution:

The work \( W \) done by a force \( \mathbf{F} \) when it acts along a displacement \( \mathbf{D} \) is given by the formula: \[ W = |\mathbf{F}| \cdot |\mathbf{D}| \cdot \cos(\theta), \] where: - \( |\mathbf{F}| \) is the magnitude of the force, - \( |\mathbf{D}| \) is the magnitude of the displacement, - \( \theta \) is the angle between the force and the displacement vectors. ### Given: - \( |\mathbf{F}| = 40 \, \text{N} \), - \( |\mathbf{D}| = 3 \, \text{m} \), - \( \theta = 60^\circ \). ### Step 1: Calculate \( \cos(60^\circ) \) We know that: \[ \cos(60^\circ) = \frac{1}{2}. \] ### Step 2: Substitute values into the work formula Now, substitute the given values into the formula for work: \[ W = 40 \times 3 \times \frac{1}{2} = 40 \times 1.5 = 60 \, \text{J}. \] ### Conclusion: The work done by the force is \( 60 \, \text{J} \). Thus, the correct answer is: \[ \boxed{60 \, \text{J}}. \]