@ elena
posted by ally .
In the figure the coefficient of static friction between mass (MA) and the table is 0.40, whereas the coefficient of kinetic friction is 0.28 ?
part a) What minimum value of (MA) will keep the system from starting to move?
part b) What value of(MA) will keep the system moving at constant speed?
physics  Elena, Tuesday, May 8, 2012 at 1:35pm
Block on the table m(A) = m1,
block on the cord m2,
the coefficient of static friction is k1=0.4,
the coefficient of kinetic friction is k2 =0.28
(a)
Block A:
T = F(fr) = k1 •N = k(s) • m1 •g,
Block B: T = m2•g.
k1 • m1 •g= m2•g,
m1 = m2/k(s) = m2/0.4.
(b)
Block A:
T = F(fr) = k2 •N = k2 • m1 •g,
Block B:
T = m2•g.
k2• m1 •g= m2•g,
m1 = m2/k2 = m2/0.28.
 thanks for the help, i am having a lot of trouble with this problem, what is m2 ?

nvm i got the answer, i was just cofused on how you explain it. Thank you . I got 5 for part a and for part b got 7.14. thank you