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A steel safe with mass 2200 kg falls onto concrete. Just
before hitting the concrete its speed is 40 m/s, and it smashes
without rebounding and ends up being 0.06 m shorter than
before. What is the approximate magnitude of the force exerted
on the safe by the concrete? How does this compare with the
gravitational force of the Earth on the safe? Explain your analysis
carefully, and justify your estimates on physical grounds.

Sagot :

The kinetic energy of the safe increases the force exerted by the concrete

to several times the weight of the safe.

  • The magnitude of the force exerted on the safe by the concrete on the is approximately [tex]\underline{29.\overline 3 \, \mathrm{MN}}[/tex]
  • The concrete exerts a force that is approximately 1,359.16 times the weight of the safe.

Reasons:

First part

The mass of the steel safe, m = 2,200 kg

Velocity of the safe just before it hits the concrete, v = 40 m/s

The amount by which the safe was compressed, d = 0.06 m

The kinetic energy, K.E., of the safe just before it hits the round is therefore;

[tex]\displaystyle K.E. = \mathbf{\frac{1}{2} \cdot m \cdot v^2}[/tex]

[tex]\displaystyle K.E._{safe} = \frac{1}{2} \times 2,200 \times 40^2 = 1,760,000 \ Joules[/tex]

Work done by concrete, W = Force, F × Distance, d

  • [tex]\displaystyle Force, \, F = \mathbf{\frac{Work, \, W}{Distance, \, d}}[/tex]

By the law of conservation of energy, we have;

The work done by the concrete, W = The kinetic energy, K.E. given by the safe

W = K.E. = 1,760,000 J

The effect of the work = The change in the height of the safe

Therefore;

The distance, d, over which the force of the concrete is exerted = The change in the height of the safe = 0.06 m

d = 0.06 m

Therefore;

[tex]\displaystyle The \ force \ of \ the \ concrete, \, F = \frac{1,760,000\, J}{0.06 \, m} = 29,333,333. \overline 3 \, N = 29.\overline 3 \ MN[/tex]

  • The force of the concrete on the safe = [tex]\underline{29.\overline 3 \ MN}[/tex]

Second part:

The gravitational force of the Earth on the safe, W = The weight of the safe

W = Mass, m × Acceleration due to gravity, g

W = 2,200 kg × 9.81 m/s² ≈ 21,582 N

The ratio of the force exerted by the concrete to the weight of the safe is found as follows;

[tex]\displaystyle Ratio \ of \ forces = \frac{29.\overline 3 \times 10^6 \, N}{21,582 \, N} = \frac{4,000,000}{2,943} \approx \mathbf{1359.16}[/tex]

  • The force exerted by the concrete is approximately 1,359.16 times the weight of the safe.

Learn more here:

https://brainly.com/question/21060171

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