A boy pushes a toy box 2.0 m along the floor by means of a force of 10 N directed downward at an angle of `60^(@)` to the horizontal.The work done by the boy is

Options

A. 6J
B. 8J
C. 10J
D. 12J

Explanation

To solve the problem of calculating the work done by the boy on the toy box, we can follow these steps:

### Step 1: Understand the Components of the Force
The boy applies a force of 10 N at an angle of 60 degrees to the horizontal. This force can be broken down into two components:
- Horizontal component: \( F_{\text{horizontal}} = F \cos(\theta) \)
- Vertical component: \( F_{\text{vertical}} = F \sin(\theta) \)

### Step 2: Calculate the Horizontal Component of the Force
Using the given values:
- \( F = 10 \, \text{N} \)
- \( \theta = 60^\circ \)

We can calculate the horizontal component:
\[ F_{\text{horizontal}} = 10 \cos(60^\circ) \]
Since \( \cos(60^\circ) = 0.5 \):
\[ F_{\text{horizontal}} = 10 \times 0.5 = 5 \, \text{N} \]

### Step 3: Determine the Displacement
The displacement of the toy box is given as:
\[ d = 2.0 \, \text{m} \]

### Step 4: Calculate the Work Done
The work done by the boy is given by the formula:
\[ W = F_{\text{horizontal}} \times d \]
Substituting the values we have:
\[ W = 5 \, \text{N} \times 2.0 \, \text{m} = 10 \, \text{J} \]

### Conclusion
The work done by the boy on the toy box is:
\[ \boxed{10 \, \text{J}} \]

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