ONLY NEED PART B. I got the -1420 for Part A ± A Sliding Crate of Fruit A crate
ID: 1594373 • Letter: O
Question
ONLY NEED PART B. I got the -1420 for Part A
± A Sliding Crate of Fruit A crate of fruit with a mass of 32.5 kg and a specific heat capacity of 3550 J/(kgK) slides 7.90 m down a ramp inclined at an angle of 36.6 degrees below the horizontal. Part A If the crate was at rest at the top of the incline and has a speed of 2.25 m/s at the bottom, how much work Wf was done on the crate by friction? Use 9.81 m/s2 for the acceleration due to gravity and express your answer in joules. Wf = -1420 J Correct Your answer 1418 J was either rounded differently or used a different number of significant figures than required for this part.
Part B If an amount of heat equal to the magnitude of the work done by friction is absorbed by the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change T? T = C SubmitHintsMy AnswersGive UpReview Part Incorrect; Try Again; 3 attempts remaining
Explanation / Answer
deltaQ = m*S*deltaT = WORKDONE BY FRICTION
where S is specific heart capacity
=>32.5*3550*deltaT = 1420
=> deltaT = 0.0123 degree celsius
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