A force of magnitude 25 N directed at an angle of `37^(@)` above the horizontal moves a 10-kg crate along a horiaontal surface at constant velocity . How much work is done by this force in moving the crate a distance of 15 m ?

Options

A. 0J
B. 40J
C. 300J
D. 98J

Answer

To solve the problem step by step, we will calculate the work done by the force acting on the crate.

### Step 1: Identify the components of the force
The force applied is 25 N at an angle of 37 degrees above the horizontal. We need to find the horizontal component of this force since work is calculated based on the component of the force in the direction of the displacement.

- The horizontal component \( F_x \) can be calculated using the cosine function:
\[ F_x = F \cdot \cos(\theta) = 25 \cdot \cos(37^\circ) \]

### Step 2: Calculate the horizontal component of the force
Using the cosine of 37 degrees, which is approximately \( \frac{4}{5} \) or 0.8:
\[ F_x = 25 \cdot \frac{4}{5} = 25 \cdot 0.8 = 20 \, \text{N} \]

### Step 3: Determine the displacement
The crate is moved a distance of 15 m along the horizontal surface.

### Step 4: Calculate the work done
Work done \( W \) is calculated using the formula:
\[ W = F_x \cdot d \]
where \( d \) is the displacement. Substituting the values we have:
\[ W = 20 \, \text{N} \cdot 15 \, \text{m} = 300 \, \text{J} \]

### Conclusion
The work done by the force in moving the crate a distance of 15 m is **300 Joules**.

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