To find the work done by the resultant force, let’s break down the components of the forces involved.

Step 1: Resolving the Force Components

The applied force $F = 30.4 \, \text{N}$ makes an angle of $20^\circ$ with the horizontal. We can resolve this force into its horizontal and vertical components.

1. Horizontal component $F_x = F \cdot \cos(20^\circ)$ $F_x = 30.4 \cdot \cos(20^\circ)$

2. Vertical component $F_y = F \cdot \sin(20^\circ)$ $F_y = 30.4 \cdot \sin(20^\circ)$

Step 2: Calculate Net Force in the Horizontal Direction

Since the box is moving horizontally, the vertical component does not contribute to the work done in this direction.

The friction force $f = 10.9 \, \text{N}$ opposes the horizontal component $F_x$. So, the resultant force in the horizontal direction $F_{\text{net}}$ is: $F_{\text{net}} = F_x - f$

Step 3: Calculate the Work Done by the Resultant Force

The work done $W$ by the resultant force is given by: $W = F_{\text{net}} \cdot d$ where $d = 11 \, \text{m}$.

Solution

1. Calculate $F_x$: $F_x = 30.4 \cdot \cos(20^\circ) \approx 30.4 \cdot 0.9397 = 28.54 \, \text{N}$

2. Calculate $F_{\text{net}}$: $F_{\text{net}} = 28.54 - 10.9 = 17.64 \, \text{N}$

3. Calculate $W$: $W = 17.64 \cdot 11 = 194.04 \, \text{J}$

Final Answer

The work done by the resultant force is approximately: $\boxed{194.04 \, \text{J}}$

Would you like more details on any step or have further questions?

Related Questions

1. What is the significance of friction in calculating work done?
2. How would the work done change if the angle of the applied force increased?
3. If the frictional force were to vary, how would we account for that in the work calculation?
4. How is the concept of work related to energy in physics?
5. What other factors could affect the work done on the box?

Tip

When calculating work done by forces at an angle, always break down forces into components to simplify calculations.