A 13 kg block(mA) rests on a level table. The coefficient of kinetic friction between the block and the table is .15. A string of negligible mass is attached to the block. The string, which is perfectly horizontal, passes through a frictionless pulley at the edge of the table and is attached to a suspended 5.0 kg block(mB)

Question

A 13 kg block(mA) rests on a level table. The coefficient of kinetic friction between the block and the table is .15. A string of negligible mass is attached to the block. The string, which is perfectly horizontal, passes through a frictionless pulley at the edge of the table and is attached to a suspended 5.0 kg block(mB)
a. Determine the acceleration of each block
b. If mA is initially at rest 1.25 m from the edge of the table, how long does it take to reach the edge of the table if the system is allowed to move freely?
c. If, instead, mB had a mass of 1.0 kg how much mass must mA contain if the acceleration of the system is kept to 1/100 g?

bobpursley · Answer

pulling force=5g
retarding force=13g*.15
total mass=18kg

Net force=totalmass*a
5g-13g*.15=(18)a solve for a

b. d=1/2 a t^2 solve for time t.

c. 1*g-mA*g=(1+mA)*g/100
100(1-mA)=1+mA
101mA=99
ma=99/101 check my math.