Edits to “Statement”, “Answer”
en/2.3.4.md
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| @@ -1,11 +1,5 @@ | |||
| ### Statement | |||
| − | $2.3.4.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| $2.3.4.$ Estimate the average force developed by a person's legs when landing after jumping from a second-floor window. | |||
| ### Solution | |||
| Let's use standard physical estimates for a human. Let the mass of the person be $m = 68\text{ kg}$. The height of a second-floor window is approximately $H = 5\text{ m}$. When landing, a person bends their knees to cushion the impact. Let the stopping distance (the vertical displacement of the center of mass during the squat) be $h = 0.68\text{ m}$. | |||
| According to the work-energy theorem, the work done by the average force of the legs goes into bringing the person's kinetic energy to zero, which originally came from the potential energy drop: | |||
| $$F \cdot h \approx mgH$$ | |||
| $$F = mg\frac{H}{h}$$ | |||
| Substituting the estimated values: | |||
| $$F = 68\text{ kg} \cdot 10\text{ }\frac{\text{m}}{\text{s}^2} \cdot \frac{5\text{ m}}{0.68\text{ m}} = 5000\text{ N} = 5\text{ kN}$$ | |||
| @@ -21,7 +15,3 @@Solution | |||
| #### Answer | |||
| $F \approx 5\text{ kN}$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $2.3.4.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $2.3.4.$ Estimate the average force developed by a person's legs when landing after jumping from a second-floor window. | $2.3.4.$ Estimate the average force developed by a person's legs when landing after jumping from a second-floor window. | ||
| ### Solution | ### Solution | ||
| Let's use standard physical estimates for a human. Let the mass of the person be $m = 68\text{ kg}$. The height of a second-floor window is approximately $H = 5\text{ m}$. When landing, a person bends their knees to cushion the impact. Let the stopping distance (the vertical displacement of the center of mass during the squat) be $h = 0.68\text{ m}$. | Let's use standard physical estimates for a human. Let the mass of the person be $m = 68\text{ kg}$. The height of a second-floor window is approximately $H = 5\text{ m}$. When landing, a person bends their knees to cushion the impact. Let the stopping distance (the vertical displacement of the center of mass during the squat) be $h = 0.68\text{ m}$. | ||
| According to the work-energy theorem, the work done by the average force of the legs goes into bringing the person's kinetic energy to zero, which originally came from the potential energy drop: | According to the work-energy theorem, the work done by the average force of the legs goes into bringing the person's kinetic energy to zero, which originally came from the potential energy drop: | ||
| $$F \cdot h \approx mgH$$ | $$F \cdot h \approx mgH$$ | ||
| $$F = mg\frac{H}{h}$$ | $$F = mg\frac{H}{h}$$ | ||
| Substituting the estimated values: | Substituting the estimated values: | ||
| $$F = 68\text{ kg} \cdot 10\text{ }\frac{\text{m}}{\text{s}^2} \cdot \frac{5\text{ m}}{0.68\text{ m}} = 5000\text{ N} = 5\text{ kN}$$ | $$F = 68\text{ kg} \cdot 10\text{ }\frac{\text{m}}{\text{s}^2} \cdot \frac{5\text{ m}}{0.68\text{ m}} = 5000\text{ N} = 5\text{ kN}$$ | ||
| @@ -21,7 +15,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $F \approx 5\text{ kN}$ | $F \approx 5\text{ kN}$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||