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

Question

4 years ago

11

ient of friction of 0.030 and a static coefficient of friction of 0.10.

2 answers:

Debora [2.8K] · Answer 1 · 2021-12-07T01:37:08+03:00

8 0

Answer:

its D 9.0

Explanation:

i got it wrong so you could get it right ;))

Anni [7] · Answer 2 · 2021-12-06T23:49:54+03:00

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