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

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

thats not a system of equations, that is a solution