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An object of mass 1000 g falls from a height of 30m on the sand below. If it penetrates 4cm into the sand, what opposing force is exerted on it by the sand? Neglect air friction.

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Final answer:

The opposing force exerted on the object by the sand, calculated using the work-energy principle and assuming no air friction, is 7350 Newtons.

Step-by-step explanation:

To determine the opposing force exerted on the object by the sand, we can use the work-energy principle. The work done by the sand to stop the object is equal to the loss of the object's kinetic energy as it penetrates into the sand.

Firstly, we calculate the object's kinetic energy just before it hits the sand. This kinetic energy comes from the gravitational potential energy (GPE) at the height of 30m. The GPE is given by the formula:

GPE = mgh

where:

  • m is the mass of the object (1000 g or 1 kg)
  • g is the acceleration due to gravity (9.80 m/s²)
  • h is the height (30 m)

Kinetic energy at the moment of impact, KE, is equal to GPE (since we neglect air friction) and given by:

KE = GPE = mgh

Next, the work W done by the sand is the force exerted F times the distance d the object penetrates:

W = Fd

Since W = KE, we can equate them to solve for F:

F = KE / d

By substituting in the values:

F = (1 kg * 9.80 m/s² * 30 m) / 0.04 m = 7350 N.

The opposing force exerted on the object by the sand is 7350 Newtons (N).

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