Potential energy and conservation of energy problem

AI Thread Summary
An outfielder throws a baseball with an initial speed of 79.0 mi/h, which converts to 35.32 m/s, while its final speed before being caught is 65 ft/s or 19.81 m/s. The problem involves calculating the mechanical energy dissipated by air drag, with the baseball's weight given as 9 oz. The kinetic energy change is calculated using the formula ΔKE=1/2mvf² - 1/2mvi², resulting in a value of -109.02 J. Clarification is sought regarding the use of "weight" versus "mass" in calculations, and there is confusion about converting units when the final answer is required in ft-lbs. The discussion emphasizes the importance of consistent unit usage throughout the problem-solving process.
magnolia1
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Homework Statement



An outfielder throws a baseball with an initial speed of 79.0 mi/h. Just before an infielder catches the ball at the same level, its speed is 65 ft/s. How much of the ball's mechanical energy is dissipated by air drag? (The weight of a baseball is 9 oz.)

Homework Equations



ke=1/2mv2
ug=mgh
Emec = ΔU + ΔK

ANSWER IS: 80.4 ft*lb

The Attempt at a Solution


Converting knowns to metres/second:
79.0 mi/h (0.44704) = 35.32 m/s
65 ft/s (0.3048) = 19.81 m/s

When they say "weight" do they mean
fg=9 oz. = 2.5 N
or
mass= 9 oz. = 0.255 kg

ΔKE=1/2mvf2-1/2mvi2
=1/2(0.255)(19.81)2-1/2(0.255)(35.32)2
=50.0356-159.0565
=-109.02 J

This is where I wasn't sure how to continue..
Any help is appreciated, thanks!
 
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hi magnolia1! :wink:
magnolia1 said:
(The weight of a baseball is 9 oz.)
…
When they say "weight" do they mean
fg=9 oz. = 2.5 N
or
mass= 9 oz. = 0.255 kg

in the ft-lb system, the pound (and the ounce) is a force

so "weight" is used instead of mass (and has the g already in it)

(the unit of mass is the slug)

anyway, why have you converted to metres etc when the required answer is in ft-lbs ?? :confused:

start again :smile:​
 

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