A body of mass 30 kg moves with an initial speed of 20 m's. If a retarding force of 60 N is applied, how long does the 2 body take to stop?

Options

A. `-10 s`
B. 10 s
C. 9 s
D. 0.10 s

Answer

To solve the problem step by step, we will use the concepts of force, mass, acceleration, and the equations of motion.

### Step 1: Identify the given values
- Mass of the body (m) = 30 kg
- Initial speed (u) = 20 m/s
- Retarding force (F) = 60 N

### Step 2: Calculate the acceleration
Using Newton's second law of motion, we know that:
\[ F = m \cdot a \]
Where:
- F is the force,
- m is the mass,
- a is the acceleration.

Since the force is a retarding force, we will consider it as negative:
\[ -60 = 30 \cdot a \]

Now, we can solve for acceleration (a):
\[ a = \frac{-60}{30} \]
\[ a = -2 \, \text{m/s}^2 \]

### Step 3: Use the equation of motion to find the time
We need to find the time (t) it takes for the body to stop. When the body stops, its final velocity (v) will be 0 m/s. We can use the equation of motion:
\[ v = u + a \cdot t \]

Substituting the known values:
\[ 0 = 20 + (-2) \cdot t \]

Rearranging the equation:
\[ 0 = 20 - 2t \]
\[ 2t = 20 \]
\[ t = \frac{20}{2} \]
\[ t = 10 \, \text{seconds} \]

### Conclusion
The body takes **10 seconds** to stop.

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