Question

Two masses of 200 grams and 400 grams are attached respectively at the end of the cord that passes through around a negligibly weight and frictionless pulley. a.) Find the acceleration of the system, b.) tension in each cord, c.) the distance and velocity of the masses when released from rest at the end of 10 seconds.

Answer 1

Tension in the cord = T
a is up for mass 1, down for mass 2

mass 1
T - .2*9.81 = .2 * a

mass 2
.4 * 9.81 - T = .4 a

solve
2 T - .4*9.81 = .4 a
-T + .4*9.81 = .4 a
--------------------subtract
3 T =- .8*9.81 = 0

T = .8 * 9.81/3

go back and get a

v = a t
d = (1/2) a t^2

Answer 2

well, sum forces.
.200g-.400g=mass(acceleration)
I took the positive direction as the direction the 200grams is headed.

-.200*9.8=(600)a
a= ....

For tension
tension: on the 200 side:
tension=.2*9.8-
.2(a)above

on the 400 side
tension=.4*9.8+.4a

c.

distance=1/2 a t^2

Answer 3

a.Find the acceleration of the system
F^1=m*a
F^2= m*g

F^1= T-F^2
400a= T-1960

F^1= F^2-T
200 a= 3920-T

200a+1960=3920-400a
a= 49/15=3.266=3.27
acceleration=3.27

b.The tension in each cord
T=200*3.27+1960
Tension=2614

T=3920-400*3.27
Tension=2612

c.The distance and velocity of the masses when released from rest at the end of 10 seconds
S=ut+1/2 at^2
S=1/2*3.27*10^2
Distance (S)=163.5m

v^2=u^2+2as
v=√(0+2*3.27*163.5)
Velocity (v)=32.7m/s^2