A 31.0 kg block is connected to anempty 1.00 kg bucket by a cord running over a

Question

A 31.0 kg block is connected to anempty 1.00 kg bucket by a cord running over a frictionless pulley.The coefficient of static friction between the table and the blockis 0.485 and the coefficient of kineticfriction between the table and the block is 0.320. Sand isgradually added to the bucket until the system just begins tomove. (a) Calculate the mass of sand added to the bucket.
(b) Calculate the acceleration of the system.
A 31.0 kg block is connected to anempty 1.00 kg bucket by a cord running over a frictionless pulley.The coefficient of static friction between the table and the blockis 0.485 and the coefficient of kineticfriction between the table and the block is 0.320. Sand isgradually added to the bucket until the system just begins tomove. (a) Calculate the mass of sand added to the bucket.
(b) Calculate the acceleration of the system.

Explanation / Answer

Mass of block m = 31 kg
mass of empty bucket m ' = 1.00 kg The coefficient of static friction between the table and theblock = 0.485 the coefficient of kinetic frictionbetween the table and the block ' = 0.320 If the static frictional force is equal toforce in forward direction then it ready to move So, Forward force ( tension ) T =mg = 147.343 N For bucket   T = ( m ' +M )g      m ' + M =15.035 So, mass of sand added M = 15.035 -m'                                          = 14.035 kg (b). When it is in motion then kineticfrictional force works So, kinetic frcrional force f = ' mg = 97.216 N for block T - f =ma               T = 97.216 + 31 a ---( 1) for bucket ( m ' + M ) g - T = ( m ' +M ) a                                       T = ( m ' + M ) ( g - a )                                          = 15.035 * ( 9.8 - a )    ----( 2) from eq ( 1 ) and ( 2 ) , 97.216 + 31a = 15.035 ( 9.8 -a )                                                        a = 1.0888 m / s ^ 2 the coefficient of kinetic frictionbetween the table and the block ' = 0.320 If the static frictional force is equal toforce in forward direction then it ready to move So, Forward force ( tension ) T =mg = 147.343 N For bucket   T = ( m ' +M )g      m ' + M =15.035 So, mass of sand added M = 15.035 -m'                                          = 14.035 kg (b). When it is in motion then kineticfrictional force works So, kinetic frcrional force f = ' mg = 97.216 N for block T - f =ma               T = 97.216 + 31 a ---( 1) for bucket ( m ' + M ) g - T = ( m ' +M ) a                                       T = ( m ' + M ) ( g - a )                                          = 15.035 * ( 9.8 - a )    ----( 2) from eq ( 1 ) and ( 2 ) , 97.216 + 31a = 15.035 ( 9.8 -a )                                                        a = 1.0888 m / s ^ 2

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