Question

a force of 10 newtons acts in a direction 75° above horizontal, moving an object 15 meters from (0, 0) to (15, 0). what is the work done by the force?
2.59 joules
3.88 joules
28.82 joules
150.00 joules

Explanation:

Step1: Recall the work formula

Work \(W=\vec{F}\cdot\vec{d}=|\vec{F}||\vec{d}|\cos\theta\)

Step2: Identify the values

\(|\vec{F}| = 10\) N, \(|\vec{d}|=15\) m, \(\theta = 75^{\circ}\)

Step3: Calculate \(\cos75^{\circ}\)

\(\cos75^{\circ}=\cos(45^{\circ} + 30^{\circ})=\cos45^{\circ}\cos30^{\circ}-\sin45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.2588\)

Step4: Calculate the work

\(W = 10\times15\times\cos75^{\circ}\approx10\times15\times0.2588 = 28.82\) J

Answer:

C. 28.82 joules