Two blocks are connected by a light string that passes over a frictionless pulle
ID: 2138484 • Letter: T
Question
Two blocks are connected by a light string that passes over a frictionless pulley as in the figure below. The system is released from rest while m2 is on the floor and m1 is a distance habove the floor.
Two blocks are connected by a light string that passes over a frictionless pulley as in the figure below. The system is released from rest while m2 is on the floor and m1 is a distance h above the floor. Assuming m1 > m2, find an expression for the speed of m1 just as it reaches the floor. (Use any variable or symbol stated above along with the following as necessary: g.) Taking m1 = 6.0 kg, m2 = 4.4 kg, and h = 3.3 m, evaluate your answer to part (a). Find the speed of each block when m1 has fallen a distance of 1.8 m.Explanation / Answer
a.let the tension in the string be T.
and common acceleration be a.
then for m1,
m1*g-T=m1*a
for m2,
T-m2*g=m2*a
solving for a,
we get
a=(m1-m2)*g/(m1+m2)
so when m1 falls through distance h,
its velocity will be =sqrt(2*acceleration*distance)=sqrt(2*(m1-m2)*g*h/(m1+m2))
(b.)
putting the values given
we get
velocity=9.95 m/s
c.)
acceleration=(6-4.4)*9.8/(6+4.4)=1.51 m/s^2
speed after falling through 1.8 m=sqrt(2*1.8*1.51)=2.33 m/s