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
![\underline{29.\overline 3 \, \mathrm{MN}}](https://tex.z-dn.net/?f=%5Cunderline%7B29.%5Coverline%203%20%5C%2C%20%5Cmathrm%7BMN%7D%7D)
- The concrete exerts a <u>force</u> that is approximately <u>1,359.16 times the weight of the safe</u>.
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;
![\displaystyle K.E. = \mathbf{\frac{1}{2} \cdot m \cdot v^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K.E.%20%3D%20%5Cmathbf%7B%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%20m%20%5Ccdot%20v%5E2%7D)
![\displaystyle K.E._{safe} = \frac{1}{2} \times 2,200 \times 40^2 = 1,760,000 \ Joules](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K.E._%7Bsafe%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%202%2C200%20%5Ctimes%2040%5E2%20%3D%201%2C760%2C000%20%5C%20Joules)
Work done by concrete, W = Force, F × Distance, d
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, <em>d</em>, 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;
![\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](https://tex.z-dn.net/?f=%5Cdisplaystyle%20The%20%5C%20force%20%5C%20of%20%5C%20the%20%5C%20concrete%2C%20%5C%2C%20F%20%3D%20%5Cfrac%7B1%2C760%2C000%5C%2C%20J%7D%7B0.06%20%5C%2C%20m%7D%20%3D%2029%2C333%2C333.%20%5Coverline%203%20%5C%2C%20N%20%3D%2029.%5Coverline%203%20%5C%20MN)
- The force of the concrete on the safe =
![\underline{29.\overline 3 \ MN}](https://tex.z-dn.net/?f=%5Cunderline%7B29.%5Coverline%203%20%5C%20MN%7D)
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;
![\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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Ratio%20%5C%20of%20%5C%20forces%20%3D%20%5Cfrac%7B29.%5Coverline%203%20%5Ctimes%2010%5E6%20%5C%2C%20N%7D%7B21%2C582%20%5C%2C%20N%7D%20%3D%20%5Cfrac%7B4%2C000%2C000%7D%7B2%2C943%7D%20%5Capprox%20%5Cmathbf%7B1359.16%7D)
- The <u>force</u> exerted by the concrete is approximately <u>1,359.16 times the weight of the safe</u>.
Learn more here:
brainly.com/question/21060171