(a) The book decelerates as it moves upwards with magnitude of 6.77 m/s²
(b) The distance traveled by the book before stopping is 0.36 m.
The given parameters;
- mass of the book, m = 1.23 kg
- mass of coffee cup, m₁ = 505 g = 0.505 kg
- initial velocity, u = 2.21 m/s
- coefficient of kinetic friction, μk = 0.221
The vertical component of the force on the book-cup system;
![F_n = mg\ cos(\theta)](https://tex.z-dn.net/?f=F_n%20%3D%20mg%5C%20cos%28%5Ctheta%29)
The frictional force on the system;
![F_k = \mu_k F_n\\\\F_k = \mu_k .\ mg\ .cos(\theta)](https://tex.z-dn.net/?f=F_k%20%3D%20%5Cmu_k%20F_n%5C%5C%5C%5CF_k%20%3D%20%5Cmu_k%20.%5C%20mg%5C%20.cos%28%5Ctheta%29)
The horizontal component of the force on the system;
![-mg\ sin(\theta) - F_k = ma\\\\-mg\ sin(\theta) -\mu_kmg \ cos(\theta) = ma\\\\-g\ sin(\theta) -\mu_kg \ cos(\theta) = a\\\\-g[sin(\theta) +\mu_k \ cos(\theta)]= a](https://tex.z-dn.net/?f=-mg%5C%20sin%28%5Ctheta%29%20-%20F_k%20%3D%20ma%5C%5C%5C%5C-mg%5C%20sin%28%5Ctheta%29%20-%5Cmu_kmg%20%5C%20cos%28%5Ctheta%29%20%3D%20ma%5C%5C%5C%5C-g%5C%20sin%28%5Ctheta%29%20-%5Cmu_kg%20%5C%20cos%28%5Ctheta%29%20%3D%20a%5C%5C%5C%5C-g%5Bsin%28%5Ctheta%29%20%2B%5Cmu_k%20%5C%20cos%28%5Ctheta%29%5D%3D%20a)
The acceleration of the book along the slope is calculated as;
![a = -g[sin(\theta) + \mu_k\ cos(\theta)]\\\\a = -9.8[sin(30) + 0.221\times cos(30)]\\\\a = -9.8(0.691)\\\\a = -6.77 \ m/s^2](https://tex.z-dn.net/?f=a%20%3D%20-g%5Bsin%28%5Ctheta%29%20%2B%20%5Cmu_k%5C%20cos%28%5Ctheta%29%5D%5C%5C%5C%5Ca%20%3D%20-9.8%5Bsin%2830%29%20%2B%200.221%5Ctimes%20cos%2830%29%5D%5C%5C%5C%5Ca%20%3D%20-9.8%280.691%29%5C%5C%5C%5Ca%20%3D%20-6.77%20%5C%20m%2Fs%5E2)
Thus, the book decelerates as it moves upwards with magnitude of 6.77 m/s²
(b) The distance traveled by the book before it comes to stop is calculated as;
![v^2 = u^2 + 2as\\\\0 = (2.21)^2 + (2)(-6.77)(s)\\\\13.54 s = 4.8841 \\\\s = \frac{4.884}{13.54} \\\\s = 0.36 \ m \](https://tex.z-dn.net/?f=v%5E2%20%3D%20u%5E2%20%2B%202as%5C%5C%5C%5C0%20%3D%20%282.21%29%5E2%20%2B%20%282%29%28-6.77%29%28s%29%5C%5C%5C%5C13.54%20s%20%3D%204.8841%20%5C%5C%5C%5Cs%20%3D%20%5Cfrac%7B4.884%7D%7B13.54%7D%20%5C%5C%5C%5Cs%20%3D%200.36%20%5C%20m%20%5C)
Thus, the distance traveled by the book before stopping is 0.36 m.
Learn more here:brainly.com/question/16037543