A force is applied for a duration of 10sec on a body of mass 5kg that is at rest. As a result the body acquires a velocity of `2ms^(-1)`. Find the magnitude of the force applied.

Answer

To find the magnitude of the force applied to the body, we can follow these steps:

### Step 1: Identify the given values
- Mass of the body (m) = 5 kg
- Initial velocity (u) = 0 m/s (the body is at rest)
- Final velocity (v) = 2 m/s
- Time duration (t) = 10 s

### Step 2: Use the first equation of motion
The first equation of motion relates initial velocity, final velocity, acceleration, and time:
\[ v = u + at \]
Where:
- \( v \) is the final velocity
- \( u \) is the initial velocity
- \( a \) is the acceleration
- \( t \) is the time duration

### Step 3: Substitute the known values into the equation
Substituting the values we have:
\[ 2 = 0 + a \cdot 10 \]

### Step 4: Solve for acceleration (a)
Rearranging the equation to solve for acceleration:
\[ a = \frac{2 - 0}{10} = \frac{2}{10} = 0.2 \, \text{m/s}^2 \]

### Step 5: Calculate the force using Newton's second law
Newton's second law states that:
\[ F = m \cdot a \]
Where:
- \( F \) is the force
- \( m \) is the mass
- \( a \) is the acceleration

### Step 6: Substitute the values for mass and acceleration
Substituting the known values:
\[ F = 5 \, \text{kg} \cdot 0.2 \, \text{m/s}^2 \]

### Step 7: Calculate the force
\[ F = 1 \, \text{N} \]

### Conclusion
The magnitude of the force applied is **1 Newton**.

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