Question

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.

Answer 1

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.