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A packet of weight w is dropped with the...

A packet of weight w is dropped with the help of a parachute and on strinking the ground comes to rest with retardation equal to twice the acceleration due to gravity. The foce exerted on the ground is

A

(a)w

B

(b)2w

C

(c)3w

D

(d)4w

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we need to analyze the forces acting on the packet when it strikes the ground and comes to rest with a specific retardation. ### Step-by-step Solution: 1. **Identify the Given Information:** - Weight of the packet = \( W \) - Retardation (deceleration) = \( 2g \) (where \( g \) is the acceleration due to gravity) 2. **Define the Forces Acting on the Packet:** - The weight of the packet acts downward and is given by \( W = mg \), where \( m \) is the mass of the packet. - The force exerted by the ground when the packet strikes it is denoted as \( F \). 3. **Calculate the Mass of the Packet:** - From the weight, we can express mass as: \[ m = \frac{W}{g} \] 4. **Apply Newton's Second Law:** - According to Newton's second law, the net force acting on the packet when it strikes the ground can be expressed as: \[ F - W = -ma \] - Here, \( a \) is the retardation, which is \( 2g \). The negative sign indicates that the acceleration is in the opposite direction to the weight. 5. **Substitute the Values:** - Substitute \( a = 2g \) and \( m = \frac{W}{g} \) into the equation: \[ F - W = -\left(\frac{W}{g}\right)(2g) \] - Simplifying the right side gives: \[ F - W = -2W \] 6. **Rearranging the Equation:** - Rearranging the equation to solve for \( F \): \[ F = W - 2W = 3W \] 7. **Conclusion:** - The force exerted on the ground by the packet when it comes to rest is: \[ F = 3W \] ### Final Answer: The force exerted on the ground is \( 3W \).
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