Which applied force will allow a 7.65 kg block of ice to begin sliding on a sheet of ice? The block of ice has a kinetic coeffic
ient of friction of 0.030 and a static coefficient of friction of 0.10.
2 answers:
Answer:
force for start moving is 7.49 N
force for moving constant velocity 2.25 N
Explanation:
given data
mass = 7.65 kg
kinetic coefficient of friction = 0.030
static coefficient of friction = 0.10
solution
we get here first weight of block of ice that is
weight of block of ice = mass × g
weight of block of ice = 7.65 × 9.8 = 74.97 N
so here Ff = Fa
so for force for start moving is
Fa = weight × static coefficient of friction
Fa = 74.97 × 0.10
Fa = 7.49 N
and
force for moving constant velocity is
Fa = weight × kinetic coefficient of friction
Fa = 74.97 × 0.030
Fa = 2.25 N
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