A constant retarding force of 20 N acts on a body of mass 5 kg moving initially with a speed of 10 `ms^(-1)`. How long does the body take to stop?

Options

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

Answer

To solve the problem, we need to determine how long it takes for a body to stop when a constant retarding force is applied. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Mass of the body (m) = 5 kg
- Initial speed (u) = 10 m/s
- Final speed (v) = 0 m/s (since the body comes to a stop)
- Retarding force (F) = 20 N

### Step 2: Calculate the acceleration
Using Newton's second law of motion, we can find the acceleration (a) caused by the retarding force. The formula is:
\[ F = m \cdot a \]
Rearranging this gives:
\[ a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{20 \, \text{N}}{5 \, \text{kg}} = 4 \, \text{m/s}^2 \]
Since this is a retarding force, the acceleration will be negative:
\[ a = -4 \, \text{m/s}^2 \]

### Step 3: Use the equation of motion to find time
We can use the first equation of motion:
\[ v = u + a \cdot t \]
Substituting the known values:
\[ 0 = 10 + (-4) \cdot t \]
This simplifies to:
\[ 0 = 10 - 4t \]
Rearranging gives:
\[ 4t = 10 \]
Thus:
\[ t = \frac{10}{4} = 2.5 \, \text{s} \]

### Final Answer
The time taken for the body to stop is **2.5 seconds**.

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