If F|= 40N (Newtons), |D| = 3m, and $\theta={60^{\circ}}$, then the work done by F acting

from P to Q is

Formula for work done:

\[ W = |F| \cdot |D| \cdot \cos\theta \]

Given:

• \( |F| = 40 \, \text{N} \)
• \( |D| = 3 \, \text{m} \)
• \( \theta = 60^\circ \)

Step 1: Plug in the values:

\[ W = 40 \cdot 3 \cdot \cos(60^\circ) \]

Step 2: Use \( \cos(60^\circ) = \frac{1}{2} \)

\[ W = 40 \cdot 3 \cdot \frac{1}{2} = 60 \, \text{J} \]

✅ Final Answer: $\boxed{60 \, \text{J}}$