a 10-kg block on a perfectly smooth horizontal table is connected by a horizontal string to a 63-kg block that is hanging over the edge of the table. What is the magnitude of the acceleration of the 10-kg block when the other block is gently released?

Answer 1

force down = m g = 63*9.81 N

mass accelerated = 63+10 = 73 kg

a = F/m = 9.81 (63/73) m/s^2

Answer 2

72

Answer 3

8.5 m/sec^2

Answer 4

by drawing both free body diagram, and using newtons second law Fnet = ma,
box 1
Ft = m1a

box 2
Ft - m2g = -m2a (because its moving down )
Ft = m2g - m2a

combining both boxes by replacing ft of box 2 in box 1 equation:
m2g - m2a = m1a
m2g = a(m1 + m2)
a = m2g / m1 + m2, use this equation and replace the number to get the acceleration value.
hope this was helpful !