A 2-oz pellet shot vertically from a spring-loaded pistol on the surface of the

Question

A 2-oz pellet shot vertically from a spring-loaded pistol on the surface of the earth rises to a height of 300 ft. The same pellet shot from the same pistol on the surface of the moon rises to a height of 1900 ft. Determine the energy dissipated by aerodynamic drag when the pellet is shot on the surface of the earth. The acceleration of gravity on the surface of the moon is 0.165 times that on the surface of the earth. (Round the final answer to three decimal places.)

The energy dissipated is ft·lb.

Explanation / Answer

Given

The Weight of pellet W = 2 0z = 2*1/16 = 1/8 lbs

1 oz = 1/16 lbs

The height to which pellet Raises on earth hE = 300 ft

The height to which pellet Raises on moon hM = 1900 ft

The accelaration of gravity in moon gM = 0.165 gE

Energy at top of flight on moon = m.hM×0.165 = .1900×0.165 = 39.188 ft.lb

Energy at top of flight on earth = m.hE = .300 = 37.5 ft.lb.

Difference = energy dissipated = 1.688 ft.lb

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