Question

1. A block of mass 57.1 kg rests on a slope having an angle of elevation of 28.3°. If pushing downhill on the block with a force just exceeding 162 N and parallel to the slope is sufficient to cause the block to start moving, find the coefficient of static friction.
HINT - Draw the block's free-body diagram and apply Newton's second law for the tilted x-component, solving for μs after substituting
fs, max = μsn.

2. A dockworker loading crates on a ship finds that a 25-kg crate, initially at rest on a horizontal surface, requires a 70-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 54 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.
static friction - ???
kinetic friction - ???

Answer 1

https://www.jiskha.com/questions/1772492/A-block-of-mass-57-1-kg-rests-on-a-slope-having-an-angle-of-elevation-of-28-3

Answer 2

What abt the second question?

Answer 3

normal force = m g = 25*9.81 both times

static 70 = static coef * mg
kinetic 54 = kinetic coef * m g

Answer 4

1 - − 0.763547
2 - Static - 0.285423
Kinetic - 0.203874

Am I right?

Answer 5

I found out that for number 2, my static coefficient is correct and my kinetic coefficient is 0.220183. Is my answer for number 1 right?

Answer 6

for number 1 I get
mg = 560
mg sin = 265
mg sin + 162 = 427
mg cos = 493
427/493 = .866

Answer 7

number 2
25 * 9.81 = 245
70/245 = .286
54/245 = .220