Asked 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?
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?
Answers
Answered by
Elena
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.
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.
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