Two blocks are connected by a light string that passes over a frictionless pulle
ID: 2178072 • 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 h above the floor.(a) 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.)
vf =
(b) Taking m1 = 6.7 kg, m2 = 3.8 kg, and h = 3.7 m, evaluate your answer to part (a).
(c) Find the speed of each block when m1 has fallen a distance of 1.5 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=4.475 m/s c. acceleration=(6.7-3.8)*9.8/(6.7+3.8)=2.706 m/s^2 speed after falling through 1.5 m=sqrt(2*1.5*2.706)=2.85 m/s