A body of mass 1 kg is initially at rest is moved by a horizontal force of 0.5 Nona smooth frictionless table. The value of work done by this force in 10 s is

Options

A. 10.9 J
B. 13.1 J
C. 14.3 J
D. 12.5 J

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the Given Information
- Mass of the body (m) = 1 kg
- Force applied (F) = 0.5 N
- Time (t) = 10 s

### Step 2: Calculate the Acceleration
Using Newton's second law of motion, we can find the acceleration (a) of the body.
\[ F = ma \implies a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{0.5 \, \text{N}}{1 \, \text{kg}} = 0.5 \, \text{m/s}^2 \]

### Step 3: Calculate the Displacement
Now, we will calculate the displacement (s) of the body after 10 seconds using the formula for displacement under constant acceleration:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the body is initially at rest, the initial velocity (u) = 0. Thus, the equation simplifies to:
\[ s = 0 + \frac{1}{2} (0.5 \, \text{m/s}^2) (10 \, \text{s})^2 \]
Calculating this:
\[ s = \frac{1}{2} \times 0.5 \times 100 = 25 \, \text{m} \]

### Step 4: Calculate the Work Done
The work done (W) by the force is given by the formula:
\[ W = F \cdot s \cdot \cos(\theta) \]
Since the force and displacement are in the same direction, \(\theta = 0^\circ\) and \(\cos(0^\circ) = 1\). Therefore:
\[ W = F \cdot s = 0.5 \, \text{N} \cdot 25 \, \text{m} = 12.5 \, \text{J} \]

### Final Answer
The work done by the force in 10 seconds is **12.5 Joules**.

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