A body of mass 1kg is dropped from a height of 5 m on to the ground. If thebody penetrates 2cm in to the groun, the average resistance offered by the ground on the body is `(g=10ms^(-2))`

Options

A. 2510 N
B. 35N
C. 50.2 N
D. 2500 N

Answer

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

### Step 1: Calculate the velocity of the body just before it hits the ground.
The formula for the velocity of a freely falling object is given by:
\[ v = \sqrt{2gh} \]
where:
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 5 \, \text{m} \) (height from which the body is dropped)

Substituting the values:
\[ v = \sqrt{2 \times 10 \times 5} \]
\[ v = \sqrt{100} \]
\[ v = 10 \, \text{m/s} \]

### Step 2: Identify the parameters for the deceleration calculation.
When the body penetrates the ground, we have:
- Initial velocity \( u = 10 \, \text{m/s} \) (just before hitting the ground)
- Final velocity \( v = 0 \, \text{m/s} \) (when the body comes to rest)
- Distance penetrated \( s = 2 \, \text{cm} = 0.02 \, \text{m} \)

### Step 3: Use the third equation of motion to find the deceleration.
The third equation of motion is:
\[ v^2 = u^2 + 2as \]
Rearranging for \( a \):
\[ a = \frac{v^2 - u^2}{2s} \]

Substituting the values:
\[ a = \frac{0^2 - (10)^2}{2 \times 0.02} \]
\[ a = \frac{-100}{0.04} \]
\[ a = -2500 \, \text{m/s}^2 \]

### Step 4: Calculate the average resistance force offered by the ground.
Using Newton's second law, the force \( F \) can be calculated as:
\[ F = m \cdot a \]
where:
- \( m = 1 \, \text{kg} \) (mass of the body)
- \( a = -2500 \, \text{m/s}^2 \) (deceleration)

Substituting the values:
\[ F = 1 \cdot (-2500) \]
\[ F = -2500 \, \text{N} \]

The negative sign indicates that the force is in the opposite direction of the motion.

### Step 5: State the average resistance offered by the ground.
The average resistance offered by the ground on the body is:
\[ 2500 \, \text{N} \]

### Summary of the Solution:
The average resistance offered by the ground on the body is **2500 N**.

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