A block is attached to a rope. The other end of the rope is then attached to a p

Question

A block is attached to a rope. The other end of the rope is then attached to a pulley (a solid disk of mass 13.0 kg and radius 39.0 cm) The pulley is then turned (about a frictionless axis through its center) so that the rope "winds up" and the block is suspended in the air. (Figure 1). Finally, the system is released from rest at the moment shown in the figure and it is observed that during the next 4.00 s the block falls 12.5 m Compute the tension in the rope while the box is falling. Express your answer with the appropriate units. Compute the mass of the box. Express your answer with the appropriate units.

Explanation / Answer

T = tension in the rope

m = mass of block

I = moment of inertia of the pulley in shape of a disc = (1/2)Mr2

r = radius of the disc/pulley = 39cm = 0.39m

M = mass of the disc/pulley = 30kg

a = acceleration of the block

= angular acceleration of the pulley

The equation of motion for the system when it is released from rest is

mg - T = ma ------- (1)

Tr = I = (1/2)Mr2

or Tr = (1/2)Mr2(a/r) ------- as a = r

or Tr = (1/2)Mra

or T = Ma/2 -------- (2)

it's given that the block falls 12.5m in 4s after the system is released from rest.

using equation S = ut + (1/2)at2, we get

12.5m = 0 + (1/2)a(4s)2

or a = 1.5625m/s2

putting a = 1.5625m/s2 in equation (2) we get:

T = Ma/2

or T = (13kg)(1.5625m/s2)/2 = 10.15625N = tension in the rope

Putting T = 10.15625N and a = 1.5625m/s2 in equation (1) we get:

mg - T = ma

or m = T/(g-a)

or m = 10.15625N/(9.8m/s2 - 1.5625m/s2) = 1.23kg = mass of the block

This concludes the answers. If you find anything lacking please let me know.. I will resolve your query without delay....

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