A 3.0 kg box is at rest on a table. The static friction coefficient us between the box and table is 0.40, and the kinetic friction coefficient per is 0.10. Then, a 10 N horizontal force is applied to the box.

While the box is at rest, the net vertical force acting on it is

∑ F = n - w = 0

where n = magnitude of the normal force (table pushing up on the box) and w = weight of the box. So we have

n = w = mg = (3.0 kg) (9.8 m/s²) = 29.4 N

The coefficient of static friction between the table and box is 0.40, so the maximum magnitude of static friction is

f = 0.40 n = 0.40 (29.4 N) = 11.76 N

which is to say, the box will not move unless a force larger than this is applied to the box.

10 N is of course smaller than 11.76 N, so the box would stay at rest, and its acceleration would remain 0 m/s².

you have to draw the persons position over there

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~