Astudent wants to determine the coefficients of static friction and kinetic friction between a box and a plank. she places the box on the plank and gradually raises one end of the plank. when the angle of inclination with the horizontal reaches 30o , the box starts to slip and slides 2.5 m down the plank in 4.0 s. calculate the coefficients of friction.

the coefficients of static friction is 0.577

the coefficients of fkinetic friction is 0.54

Step-by-step explanation:

given information

θ = 30°

S = 2.5 m

t = 4 s

- Static friction

F - F(friction) = 0

W sin θ = μs W cos θ  

μs = sin θ/ cosθ

   = tan θ

   = tan 30°

   = 0.577

- kinetic friction

S = +

= 0

S =

a = =

  = 0.3

F - F(friction) = m a

W sin θ - μ W cos θ  = m a

mg sin θ - μk mg cos θ  = m a

g sin θ - μk g cos θ  = a

μk = g sin θ  - a / g cos θ

    = 9.8 sin 30° - 0.3 / 9.8 cos 30°

    = 0.54

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116