Work and Kinetic Energy of baseball

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To calculate the work required to throw a baseball at 90 mph, the equation W = (change) KE, where KE = 1/2 mv^2, is used. The baseball weighs 5 oz, which needs to be converted correctly to mass in slugs for accurate calculations. The conversion from miles per hour to feet per second is also necessary. The correct answer, as stated in the calculus book, is 85.1 ft-lbs, which highlights the importance of proper unit conversions in physics problems. The initial confusion was resolved by converting the weight of the ball to its mass using gravitational acceleration.
saraleigh117
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How many foot pounds of work does it take to throw a baseball 90 mph? A baseball weighs 5 oz, or 0.3125 lb.

The only equation I can find is W= (change) KE, where KE=1/2mv^2

When I try plugging in the data given I'm not getting the correct answer. I don't know if I'm doing incorrect unit conversions or what. This is actually in my calculus book physics but this seemed like the more appropriate place to post the question. By the way, the answer in the book is 85.1 ft-lbs. Please help me. I'm dyin' over here.
 
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You have the right equation, so you must be doing your conversions incorrectly, which is a little tough in the USA units when mass comes into play. You must first convert miles/hr to ft/sec, and pounds to slugs per m=W/g.
 
Post your calculations.
 
Nevermind, I've got it. I didn't convert the weight of the ball to the mass of the ball using the gravitational acceleration. Thanks.
 

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