a 14.0 kg mass, m_1, on a frictionless table is accelerated by an 8.0 kg mass, m_2, hanging over the edge of?

Question

the table. what is the acceleration of the mass along the table?

Answer 1

a=3.56 m/s^2

for 14kg diagram

D.........> T ( point to the right of the table cuz 8kg try to fall down )
sum of the force on x direction is T= ma = 14a

( i called D is the block 14kg, T is tension of string)

for the block 8kg
.
.
. T going up
.
D ( block 8kg)
.
.mg is pulling down

we choose the y + is going down cuz acceleration is going down
sum of the force in y direction is
mg- T = ma or 8(9.8) -T = 8a
or 78.4 -T = 8a

but T = 14a above substitute into 78.4 -T = 8a
then 78.4 -14a = 8a
add 14a both sides
78.4= 22a
a= 78.4/22 = 3.56m/s^2

Answer 2

The weight in Newton applied by the hanging mass from the edge of the frictionless table = 8*9.8= 78.4 N

Assuming no friction, or change of cable tension at the edge of the table, then F= ma; 78.4 = 14*a
a = 78.4/14= 5.6 m/s^2

Answer 3

lonelyspirit is mistaken. He forgot that the 78.4N force is accelerating TWO masses. The correct equation is:

78.4N = (14kg + 8kg)(a)

So the acceleration comes out to 3.56 m/s²