Two masses are connected by a light string passing over a light, frictionless pulley.?

Question

The m1 = 5.35 kg object is released from rest at a point 4.00 m above the floor, where the m2 = 3.30 kg object rests.


(a) Determine the speed of each object when the two pass each other.
m/s
(b) Determine the speed of each object at the moment the 5.35 kg mass hits the floor.
m/s
(c) How much higher does the 3.30 kg mass travel after the 5.35 kg mass hits the floor?
m

Answer 1

Das zounds az Ahtung mashine!
No it is Atwood machine

Zis iz wat I said Atswud mashin!

F=W1 - W2 (no inertia or weight of the string)
(m1+m2)a=(m1-m2)g

a=g(m1-m2)/(m1+m2)

a)They will pass each other with speed
V=at
when
h - distance passed by m1 on the way down
H-h - distance passed by m2 on the way up where H=4.00m

H-h=h or h=(1/2)H
h=(1/2)at^2
t=sqrt(2h/a) or
t=sqrt(H/a)
V=at so finally
V= g[(m1-m2)/(m1+m2)] sqrt(H/a)
b) Similarly to a accept t=sqrt(2H/a)
or use energy relationship

Pe1-Pe2=Ke1+Ke2
g(m1-m2)H=(1/2) (m1+m2)V^2
V=sqrt(2g(m1-m2)H/(m1+m2)) (you should get the same equation)
c) Ke=Pe
(1/2) m2 V^2=m2 g y
y=V^2/(2 g )

Answer 2

Start with drawing a picture and making free body diagrams for both crates. We know that sum of all forces = ma, so Crate 1: m2a = T - Fg2sin(theta) Fg = force due to gravity and T = tension and Crate 2: m1a = Fg1 - T solve for the unknown values. I started with T = Fg1 - m1a (m1 = mass of m1). Plug this into the other equation so m2a = (Fg1 - m1a) - Fg2sin(theta) (6kg)a = (98N - 10kg(a)) - 58.8Nsin(39 degrees) a is approximately = 3.81 m/s^2 Plug this value into an equation to find T. (10kg)(3.81m/s^2) = 98N - T T is approximately 59.9N