Asked by Chris
The 3.25 kg physics book shown is connected by a string to a 472.0 g coffee cup. The book is given a push up the slope and released with a speed of 3.79 m/s. The coefficients of friction are μs = 0.595 and μk = 0.241. What is the acceleration of the book if the slope is inclined at 29.5°?
MY ATTEMPT:
Assume x is parallel with slope. Assume positive direction is direction of Vi.
-Fnet(x) = F(wsin(theta)) + F(fk) + F(cup)
F(wsin(theta)) = (9.81*3.25)sin29.5 = 15.6997 N
F(fk) =uK* Fn = (0.241)((9.81*3.25)cos29.5) = 6.6875 N
F(cup) = w(cup) = 9.81* 0.472kg = 4.6303 N
-Fnet(x) = 15.6997 N + 6.6875 N + 4.6303 N = -27.018 N
F = ma
a = F/m = -27.018 N / 3.25 kg = -8.31 m/s^2
However this is the wrong answer. Can anyone see what I'm doing wrong?
Thank you very much.
MY ATTEMPT:
Assume x is parallel with slope. Assume positive direction is direction of Vi.
-Fnet(x) = F(wsin(theta)) + F(fk) + F(cup)
F(wsin(theta)) = (9.81*3.25)sin29.5 = 15.6997 N
F(fk) =uK* Fn = (0.241)((9.81*3.25)cos29.5) = 6.6875 N
F(cup) = w(cup) = 9.81* 0.472kg = 4.6303 N
-Fnet(x) = 15.6997 N + 6.6875 N + 4.6303 N = -27.018 N
F = ma
a = F/m = -27.018 N / 3.25 kg = -8.31 m/s^2
However this is the wrong answer. Can anyone see what I'm doing wrong?
Thank you very much.
Answers
Answered by
drwls
The term F(cup) is the tension in the string connecting cup and book. This does NOT equal M(cup)*g because the cup is accelerating downward at a rate less than g.
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