Question

Two blocks start from rest and are connected by a string that passes over a pulley, as shown in the picture below. The pulley is a uniform disk of mass
mp = 4.00 kg; m = 5.00 kg and M = 11.0 kg. For both blocks, ?k = 0.240. If the speed of the blocks is 1.50 m/s after they have moved a distance of 2.00 m, find ?.

Can you please give a detailed explanation of how you got the answer including any free body diagrams and formulas you used. The answer is 24.9°

Answer 1

the acceleration of the system is

$a = \frac{v^2}{2S} = \frac{1.5^2}{2(2)} = 0.5625m/s^2$

The equation of motion for inclined plane object

$M g \sin \theta - \mu_k M g \cos \theta = M a$

$(107.8) \sin \theta - (25.872) \cos \theta - T_1= 6.1875$ -----(1)

on the top

$T_2 - \mu_k m g = m a$

----(2)

Now for the pulley

$(T_1- T_2) r = I \alpha$

$(T_1- T_2) r = \frac{1}{2}m_pr^2 \left ( \frac{a}{r} \right )$

$T_1- T_2 = \frac{1}{2}m_p a$

$T_1- T_2 = \frac{1}{2}(4) (0.5625)$

$T_1 = \frac{1}{2}(4) (0.5625) + T_2$

$T_1 = \frac{1}{2}(4) (0.5625) + 14.5725\Rightarrow 15.6975 N$ ----(3)

put this equation in (1) for angle

$(107.8) \sin \theta - (25.872) \cos \theta - 15.6975= 6.1875$

$(107.8) \sin \theta - (25.872) \cos \theta = 21.885$

$(107.8) \sin \theta - (25.872) \sqrt{ (1 - \sin^2 \theta )}= 21.885$

squaring the equations on either sides and solving for the angle

$\theta = 24.885^o$    (from wolframe alpha solver )