Question

A friction-less pulley in the shape of a solid cylinder of mass 2.90 kg and radius 30.0 cm is used to draw water from a well. A bucket of mass 1.90 kg is attached to a cord wrapped around the cylinder. If the bucket starts from rest at the top of the well and falls for 3.70 seconds before hitting the water, how deep is the well

Answer 1

Answer:

$s=-38.332m$

Explanation:

From the question we are told that

Cylinder of mass $m=2.90 kg$

Radius $R=30.0 m$

Mass of bucket $M_b= 1.90 kg$

Fall time $T_f= 3.70 seconds$

Generally the tension the bucket is mathematically given by

$T-mg=ma$

For cylinder

 $T=I*\alpha$

 where

$I=1/2mR^2$

$a=R*\alpha$

Giving

$T=-1/2mR\alpha$

Therefore

$a=\frac{-mg}{m+1/2m}$

$a=\frac{-(1.9*9.81)}{1.9+0.5*(2.9)}$

$a=-5.563880597 \approx -5.6m/s^2$

Generally the Newton's equation for motion is mathematically represented as

 $s=ut+1/2at^2$

 $s=0(3)+1/2*-5.6*(3.7)^2$

$s=-38.332m$