Consider the system shown in the figure (Figure 1) . Block A weighs 43.3N and bl
ID: 1285892 • Letter: C
Question
Consider the system shown in the figure (Figure 1) . Block A weighs 43.3N and block B weighs 19.0N . Once block B is set into downward motion, it descends at a constant speed
Calculate the coefficient of kinetic friction between block A and the tabletop.
A cat, also of weight 43.3N , falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration magnitude?
A cat, also of weight 43.3N , falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration direction?
Consider the system shown in the figure (Figure 1) . Block A weighs 43.3N and block B weighs 19.0N . Once block B is set into downward motion, it descends at a constant speed Calculate the coefficient of kinetic friction between block A and the tabletop. A cat, also of weight 43.3N , falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration magnitude? A cat, also of weight 43.3N , falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration direction?Explanation / Answer
f = T
u*Wa = Wb
u = Wb/Wa = 19/43.3 = 0.44
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wb* - T = ma*a============> T = wb - mb*a
T - u*(wa+wc) = (ma+mc)*a
wb - mb*a - u*(wa+wc)= (ma+mc)*a
a = (wb - u*(wa+wc))/(ma+mb+mc))
a = g*(wb - u*(wa+wc))/(wa+wb+wc)
a = - 1.78m/s^2
magnitude = 1.78 m/s^2
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(wb+wc)- T = (mb+mc)*a============> T = (wb+wc)- (mb+mc)*a
T - u*(wa) = ma*a
(wb+wc)- (mb+mc)*a - u*(wa)= ma*a
a = (wb+wc - u*(wa))/(ma+mb+mc)
a = g*(wb+wc - u*(wa))/(wa+wb+wc)
a = 4.01 m/s^2