Question

If a force of $(6\sqrt{3}\hat{i}-4\sqrt{2}\hat{j})\text{ N}$ acts on a body of mass 3.7 kg and displaces the body by $(2\sqrt{3}\hat{i})\text{ m}$, then the work done by the force is:}

• A. 18 J
• B. 40 J
• C. 36 J
• D. 52 J

Hint:

Only components in same direction contribute in dot product.

Answer 1

The Correct Option is C

Step 1: Work done is given by dot product: \[ W = \vec{F}\cdot \vec{d} \]
Step 2: Given force: \[ \vec{F} = 6\sqrt{3}\hat{i} - 4\sqrt{2}\hat{j} \] Displacement: \[ \vec{d} = 2\sqrt{3}\hat{i} \]
Step 3: Only i-component contributes since displacement has no j-component.
Step 4: Compute dot product: \[ W = (6\sqrt{3})(2\sqrt{3}) + (-4\sqrt{2})(0) \]
Step 5: \[ W = 12 \times 3 = 36\ \text{J} \]