Static and Kinetic Friction on an incline: A 10 kg crate is placed at rest on an

Question

Static and Kinetic Friction on an incline:
A 10 kg crate is placed at rest on an incline plane (s =0.8, k =0.4), whose angle can be adjusted.

a.) The maximum angle of the incline for which the crate will stay at rest is:
degrees

b.)-e.) If the angle of the incline were instead the two amounts shown below, please fill in the rest of the values on the chart below:
angle of incline: component of weight acting parallel to incline: (size only): 20 degree, 60 degree
friction force: (size only): N, N
(indicate static or kinetic):
net force on crate:
(size only)
acceleration of crate: M/S^2
(size only)
20o N N, N m/s2
60o N N, N m/s2


f.) For the 60o angle, find how long it would take the crate to slide down 5 m along the incline.
s

Explanation / Answer

(a) let at angle , the crate is at rest.

at maximum value of for which the crate is at rest,

(component of weight parallel to incline)=(static friction force)

=> mg sin=sN

                =smg cos

=> tan=s=0.8

=>max=tan-1(0.8)=38.680

(b)at 200, crate will be at rest, as <38.680

thus, component of weight acting parallel to incline= mg sin=10*9.8*sin(20)=33.50 N

friction force=component of weight acting parallel to incline=33.50 N [static]

net force on the crate=0 [as crate is at rest]

acceleration of the crate=0

at 600, the crate will slide down as >38.680

component of weight acting parallel to incline = mg sin=10*9.8*sin(60)=84.84 N

friction force= kinetic friction=k mg cos=0.4*10*9.8*cos(600)=19.62 N

net force=84.84-19.62=63.22 N

acceleration=net force/mass=63.22/10=6.322 m/s2

(c)from S=ut+(0.5)at2

here, S= 5m; u=0; a=6.322 m/s2

5=(0.5)*6.322*t2

=>t=1.26 s

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