An object whose mass is 300 lb experiences changes in its kinetic and potential energies owing to the action of a resultant force r. the work done on the object by the resultant force is 140 btu. there are no other interactions between the object and its surroundings. if the object's elevation increases by

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

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