a constant force F=(m2g)/2 is applied on the block of mass m1?as shown in figure. the string and pulley are light and surface of the table is smooth. find the acceleration of m1

F - T(tension)=m1a
T-m2g=m2a
add both
F- m2g= m1a=m2a
F-m2g=(m1+m2)a
F=m2g/2[given]
m2g/2-m2g=(m1+m2)a
-m2g/2=(m1+m2)a
a = -m2g/2(m1+m2)

35

Raushan Kumar answered this

A constant force F = m2
g/2 is applied on the block
of mass m1
as shown in fig. The string and the
pulley are light and the surface of the table is
smooth. The acceleration of m1
is –

0

Goswami Dharmeshgiri Ghanshyam... answered this

A constant force
=
2
is applied on the block of mass
1
as shown in the figure. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of
1
.