A force of 5 N, making an angle `theta` with the horizontal, actig on an object displaces it by 0.4 m along the horizontal direction. If the object gains kinetic energy of 1J. The horizontal component of the force is

Options

A. 1.5N
B. 2.5 N
C. 3.5 N
D. 4.5 N

Answer

To solve the problem step by step, we will use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Force \( F = 5 \, \text{N} \)
- Displacement \( d = 0.4 \, \text{m} \)
- Change in kinetic energy \( \Delta KE = 1 \, \text{J} \)

2. **Understand the Work-Energy Theorem:**
- According to the work-energy theorem, the work done \( W \) on an object is equal to the change in kinetic energy:
\[ W = \Delta KE \]
- Therefore, we can write:
\[ W = 1 \, \text{J} \]

3. **Calculate the Work Done:**
- The work done by the horizontal component of the force \( F_h \) is given by:
\[ W = F_h \cdot d \]
- Since the displacement is in the horizontal direction, we can express the work done as:
\[ 1 \, \text{J} = F_h \cdot 0.4 \, \text{m} \]

4. **Solve for the Horizontal Component of the Force:**
- Rearranging the equation to find \( F_h \):
\[ F_h = \frac{1 \, \text{J}}{0.4 \, \text{m}} = 2.5 \, \text{N} \]

5. **Conclusion:**
- The horizontal component of the force \( F_h \) is \( 2.5 \, \text{N} \).