Answer :
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.
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