A 8.50 kg block (m1) and a 7.55 kg block (m2) rest on a horizontal frictionless surface with m2 resting on top of m1. The surface between the top and bottom blocks is roughened so that there is no slipping between the two blocks. A 42.6 N force is applied to the bottom block.
What is the acceleration for the two block system?
What is the force of static friction between the top and bottom blocks?
What is the minimum coefficient of static friction necessary to keep the top block from slipping on the bottom block?
2006-10-09 07:58:59 · 1 answers · asked by gunder53534 2 in Science & Mathematics ➔ Physics
total force = (m1 +m2)*a
42.6=16.05*a
a =2.65 units
if it is not sliiping then definitely it might be equal to or greater than
42.6 N
u *m2* g=42.6
u=0.575(coff. of static friction)
2006-10-09 08:31:04 · answer #1 · answered by n nitant 3 · 0⤊ 0⤋