A crate weighing 50 N is being pulled 10 meters across the horizontal floor by a 30-N force applied horizontally against 20 N of friction. What’s the work done by the applied force? Assume gravity is 10m/s^2

To calculate the work done by the applied force, we use the formula:

\[
W = F \cdot d \cdot \cos\theta
\]

Where:
- \(W\) is the work done,
- \(F\) is the force applied,
- \(d\) is the distance over which the force is applied,
- \(\theta\) is the angle between the force and the direction of motion.

### Given:
- \(F = 30 \, \text{N}\) (applied force),
- \(d = 10 \, \text{m}\),
- \(\theta = 0^\circ\) (force is applied horizontally, so \(\cos\theta = \cos 0^\circ = 1\)).

The formula simplifies to:

\[
W = F \cdot d \cdot \cos 0^\circ = F \cdot d
\]

Substituting the values:

\[
W = 30 \, \text{N} \cdot 10 \, \text{m} = 300 \, \text{J}
\]

### Result:
The work done by the applied force is **300 joules (J)**.

Let me know if you'd like additional details or have questions. 

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Here are 5 related questions to explore:
1. What would the work be if the force were applied at an angle of \(30^\circ\)?
2. How much work is done by the frictional force in this scenario?
3. What is the net work done on the crate?
4. How would the work change if the friction force increased to 25 N?
5. What is the acceleration of the crate given the net force?

**Tip:** Always consider the angle of application when calculating work, as it can significantly affect the result.

A crate weighing 50 N is being pulled 10 meters across the horizontal floor by a 30-N force applied horizontally against 20 N of friction. What’s the work done by the applied force? Assume gravity is 10 m/s^2.

Understand how to calculate the work done by a 30-N horizontal force pulling a crate 10 meters across a floor with 20 N of friction. Learn key physics concepts such as work, forces, and the impact of friction. Suitable for high school students in Grades 9-12.

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