A thief stole a box full of valuable articles of weight W and while carrying it on his back, he jumped down a wall of height ‘ h ’ from the ground. Before he reached the ground he experienced a load of

Options

A. 2 W
B. `W`
C. `W//2`
D. Zero

Answer

To solve the problem, we need to analyze the situation when the thief jumps down from a height 'h' while carrying a box of weight 'W'. We will determine the load experienced by the thief just before he reaches the ground.

### Step 1: Understand the forces acting on the thief
When the thief jumps down, two main forces act on him:
1. The gravitational force acting downwards due to his weight and the weight of the box, which is \( W \).
2. The force of impact when he lands, which will momentarily increase the load he feels.

### Step 2: Calculate the gravitational force
The total weight that the thief is carrying is simply the weight of the box, which is given as \( W \). Therefore, the total downward force due to gravity when he is in free fall is:
\[ F_{\text{gravity}} = W \]

### Step 3: Analyze the motion during the fall
When the thief jumps down from a height 'h', he will accelerate downwards due to gravity. The acceleration due to gravity is denoted as \( g \) (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 4: Determine the velocity just before impact
Using the kinematic equation for free fall:
\[ v^2 = u^2 + 2gh \]
where \( u = 0 \) (initial velocity when he jumps), we find:
\[ v^2 = 0 + 2gh \]
Thus, the velocity just before impact is:
\[ v = \sqrt{2gh} \]

### Step 5: Calculate the load experienced upon landing
When the thief lands, he experiences an additional force due to the deceleration as he comes to a stop. The load experienced can be calculated using the impulse-momentum principle or by considering the deceleration.

Assuming he comes to a stop over a very short distance \( d \) (the distance over which he decelerates), the average force experienced can be approximated as:
\[ F_{\text{impact}} = W + \Delta F \]
where \( \Delta F \) is the additional force experienced due to the deceleration.

Using Newton's second law, the additional force can be expressed as:
\[ \Delta F = m \cdot a \]
where \( a \) is the deceleration. The deceleration can be calculated from the change in velocity over the time taken to stop.

### Step 6: Conclusion
The load experienced by the thief just before he reaches the ground can be expressed as:
\[ F_{\text{total}} = W + \Delta F \]

This means that the total load experienced by the thief just before he reaches the ground is greater than the weight \( W \) due to the impact force.