Question

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

• A. \( 60\sqrt{3} \) J
• B. \( 120 \) J
• C. \( 60\sqrt{2} \) J
• D. \( 60 \) J

Hint:

Work done is a scalar quantity.

Answer 1

The Correct Option is D

Step 1: Identify the given information.
Magnitude of the force, \( |\vec{F}| = 40 \) N
Magnitude of the displacement, \( |\vec{D}| = 3 \) m
Angle between the force and displacement vectors, \( \theta = 60^{\circ} \)
Step 2: Recall the formula for work done. \[ W = |\vec{F}| |\vec{D}| \cos \theta \] Step 3: Substitute the given values into the work formula. \[ W = (40) (3) \cos(60^{\circ}) \] Step 4: Evaluate the cosine of the angle. \[ \cos(60^{\circ}) = \frac{1}{2} \] Step 5: Calculate the work done. \[ W = 40 \times 3 \times \frac{1}{2} = 120 \times \frac{1}{2} = 60 { J} \]