± A Sliding Crate of Fruit A crate of fruit with a mass of 34.5 kg and a specifi
ID: 1510439 • Letter: #
Question
± A Sliding Crate of Fruit
A crate of fruit with a mass of 34.5 kg and a specific heat capacity of 3500 J/(kgK) slides 7.70 mdown a ramp inclined at an angle of 38.0 degreesbelow the horizontal.
Part A
If the crate was at rest at the top of the incline and has a speed of 2.50 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.
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Correct
The frictional force opposes the motion of the crate, so the work done on the crate by friction must be a negative quantity.
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?
± A Sliding Crate of Fruit
A crate of fruit with a mass of 34.5 kg and a specific heat capacity of 3500 J/(kgK) slides 7.70 mdown a ramp inclined at an angle of 38.0 degreesbelow the horizontal.
Part A
If the crate was at rest at the top of the incline and has a speed of 2.50 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 = -1500 JSubmitHintsMy AnswersGive UpReview Part
Correct
The frictional force opposes the motion of the crate, so the work done on the crate by friction must be a negative quantity.
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?
Explanation / Answer
potential energy at the top Ui = m*g*L*sintheta
kinetic energy at the top = Ki = 0
total energy at the top = Ei = ui + ki = m*g*L*sintheta
potential energy at the bottom Uf = 0
kinetic energy at the bottom = Kf = 0.5*m*v^2
total energy at the bottom = Ef = uf + kf = 0.5*m*v^2
from work energy theorem
fWf = Ei - Ei
Wf = (0.5*34.5*2.5^2)-(34.5*9.8*7.7*sin38) = -1500 J
+++++++++++++++++++++++
amount of heat gained Q = m*C*dT
Q = Wf
m*C*dT = 1500
34.5*3500*dT = 1500
dT = 0.0124 degrees
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