A body of mass 2kg is dropped from a height of 4m. If the average resistance offered by the ground is 4020N, the distance through which the body penetrates in to the ground before coming to rest is (g = 10ms^–2)

Step 1: Calculate the Potential Energy at the Starting Height

The body initially has potential energy due to its height from the ground. Using the formula:

PE = mgh

Where:

• m = 2 kg (mass of the body)
• g = 10 m/s² (acceleration due to gravity)
• h = 4 m (height)

Substituting the values:

PE = 2 × 10 × 4 = 80 J

Thus, the potential energy of the body at height 4m is 80 J.

Step 2: Energy Conservation

When the body of mass 2kg falls, its potential energy is completely converted into kinetic energy (KE) just before hitting the ground:

KE = PE = 80 J

This energy is then used to overcome the resistance force of the ground as the body penetrates it.

Step 3: Work Done Against the Ground Resistance

The work done by the ground resistance (F) over the penetration distance (S) is equal to the kinetic energy of the body:

F × S = KE

Where:

• F = 4020 N (average resistance force)
• S = ? (penetration distance)
• KE = 80 J (total energy)

Substituting the values:

4020 × S = 80

Step 4: Solve for Penetration Distance

Rearranging the equation to solve for S:

S = 80 / 4020 = 0.02 m

Convert meters to centimeters:

S = 0.02 m × 100 = 2 cm

Thus, the body of mass 2kg penetrates the ground by a distance of 2 cm before coming to rest.

Final Answer

The distance through which the body of mass 2kg penetrates into the ground is 2 cm.

This calculation relies on the principle of energy conservation, where the body’s potential energy is converted to work done against the ground resistance.