Answer to Question #95736 in Mechanics | Relativity for stefanie

Question

Consider the Atwood machine shown in the figure, where
m1 = 2.00 kg
and
m2 = 5.90 kg.
The system starts at rest, then the sphere is given a quick push downward, giving it an initial speed of 2.95 m/s.
Assume the pulley and cord are massless, and the cord is inextensible. Neglect friction.
a. Through what distance (in m) will m1 descend?
b.What is the velocity (in m/s) of m1 after 1.80 s?
magnitude and direction

Answer 1

Since the mass m2 is heavier than m1, both masses experience the same but oppositely directed acceleration. Calculate that acceleration which will be upward for m1 and downward for m2 (y-axis up) by solving the system of equations below written according to Newton's second law:

$$T-m_2g=-m_2a,\\ T-m_1g=m_1a.$$

Therefore:

$$a=4.84\text{ m/s}.$$

a. m1 will descend for h meters, then stop, and then it will start accelerating upward. Use common equations from kinematics to find the distance:

$$h=\frac{v_0^2}{2a}=0.90\text{ m}.$$

b. The velocity of m1 after 1.80 s is

$$u=-v_0+at=5.762\text{ m/s}.$$

Since we pointed y-axis upward, the initial velocity was directed downward and the acceleration upward, the velocity after 1.8 s will be directed upward.