Answer: initial speed = 151.89 ft/s
Explanation:
1) About the question:
The question is incomplete. The final missing part is: If the object's elavation increases by 100 ft and its final velocirty is 200 ft/s, what is its initial velocity, in ft/s? Let g= 32.2 ft/s^2
2) Physical principle:
Work = change in mechanical energy
3) Formulas
Kinetic energy, KE = (1/2)mv²
Potential energy, PE = mgh
Mechanical energy, ME = KE + PE
4) Conversion of units:
i) 300 lbs = 300/2.2046 kg = 136.08 kg
ii) 140 btu = 140 × 1055.06 J = 147,708.4 J
iii) g = 9.81 m/s²
iv) 100 ft = 100 × 0.3048m = 30.48 m
v) 200 ft/s = 200 × 0.3048 m/s = 60.96 m/s
5) Solution
i) change in PE:
ΔPE = mgΔh = 136.08kg × 9.81 m/s² × 30.48m = 40,689.12 J
ii) change in KE
ΔKE = work done on the object - ΔPE = 147,708.4J - 40,689.12J = 107,019.28 J
iii) initial KE
initial KE = final KE - ΔKE = (1/2)mv² - ΔKE = (1/2) (136.08kg)(60.96m/s)² = 252,844.91J - 107,019.28J = 145,825.63J
iv) initial speed
KE = (1/2)mv² ⇒ v² = 2KE / m = 2 (145,825.63J) / 136.08 kg = 2,143.23 m²/s²
v = √v² = 46.30 m/s
v) Convert to ft/s
46.30 m/s = 46.30 / 0.3048 ft/s = 151.89 ft/s
