How much work is done to slide a crate for a distance of $25$ m along a loading dock by pulling on it with a $180$ N force where the dock is at an angle of $45^\circ$ from the horizontal?
- A. $3.18198 \times 10^3$ J
- B. $3.18198 \times 10^2$ J
- C. $3.4341 \times 10^3$ J
- D. $3.4341 \times 10^4$ J
✅ Correct Answer: $3.18198 \times 10^3$ J
Explanation
Work done is given by $W = F d \cos \theta$ Here $F = 180$ N, $d = 25$ m and $\theta = 45^\circ$ $W = 180 \times 25 \times \cos 45^\circ$ $= 180 \times 25 \times \frac{1}{\sqrt{2}}$ $= 3.18198 \times 10^3$ J
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