A12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.00-kg weight by a thin, light wire that passes over a frictionless pulley (fig. e10.16).

the pulley has the shape of a uniform solid disk of mass 2.10 kg and diameter 0.560 m.

after the system is released, find:

(a) the tension in the wire on both sides of the pulley

(b) the acceleration of the box

(c) the horizontal and vertical components of the force that the axle exerts on the pulley.

Explanation:

a )

Since pulley is friction-less , tension on both the sides of pulley will be same.

Let it be T . The pulley will not rotate because it is friction-less.

For motion of 5 kg weight in downward direction with acceleration a

mg - T = ma

For motion of box  on horizontal surface

T = Ma ( M is mass of box )

mg - Ma = ma

a = mg / (m +M)

= 5 x 9.8 / 17

= 2.88 m / s

T = Ma

= 12 x 2.88

= 34.55 N

c )

Horizontal component of the force that the axle exerts on the pulley

= Tension in the wire

= 34,55 N

Vertical  component of the force that the axle exerts on the pulley

= Tension in the wire

= 34,55 N

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be