Question

In the Atwood machine shown in Figure 5.14a, m1 = 2.00 kg and m2 = 8.00 kg. The masses of the pulley and string are negligible by comparison. The pulley turns withoutfriction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 2.00 m/s downward.

Figure 5.14a
(a) How far will m1 descend below its initial level?
_______m
(b) Find the velocity of m1 after 1.80 s.
_______m/s

Answer 1

vf^2=vo^2 +2a(yf-yo) 

 

where a= (m2-m1/m2+m1)(g)=(7.5-2)/(7.5+2)g=11g/19 

 

0=(-2.4)^2 + 2(11g/19)(yf-0) 

 

a.) yf=-.507 m below its initial level 

 

yf-yo = 1/2*(vo+vf)*(tf-to), to=0, yo=0 

 

-.507 = 1/2*(-2.4 + 0)*tf 

 

tf=.423 sec, so after .423 seconds, the mass m1, stops moving downward, and begins to accelerate upward 

 

1.8 - .423 = 1.377s, so for 1.377 the mass, m1, accelerates upward, m2 accelerates downward at same rate, where m1 achieves a final velocity upwards of 

 

vf = vo + a(tf-to) 

vf = 0 + (11g/19)*1.377 = 

 

b.) 7.81 m/s upward 

 

note: this assumes there is enough string that neither m1 nor m2 crash into the pulley. A reasonable assumption since the string length is not given and acceleration is independent of string length, it only depends on the masses m1 and m2, and g.