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.

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.

To solve this problem, you seem to have used the correct formulas and values for calculating the net force and acceleration. However, there is a small error in your calculation of the frictional force (Ffk).

The frictional force is given by Ffk = μk * Fn, where Fn is the normal force. In this case, the normal force is equal to the weight of the book component perpendicular to the slope, which is Fw * cos(theta). Here, Fw is the weight of the book, given by Fw = m * g = 3.25 kg * 9.81 m/s^2.

So, the correct calculation for the frictional force becomes:

Ffk = μk * Fn = μk * Fw * cos(theta)
= 0.241 * (3.25 kg * 9.81 m/s^2) * cos(29.5°)

Make sure to convert the angle from degrees to radians before calculating the cosine.

Once you have the correct value for Ffk, recalculate the net force and the acceleration using the corrected values. Your final answer should match the correct solution.