A box slides with uniform acceleration up an incline. The box has an initial speed of 9.0 m/s and rises vertically 2.60 m before
coming to rest. If the angle of the incline is 30°, what is the coefficient of kinetic friction between the box and the incline?
1 answer:
Answer: 0.58
Explanation:
First we need to get the acceleration of the body using equation of motion
v²=u²-2as
v is the final velocity
u is the initial velocity
a is acceleration
s is the distance moved
0²=9²-2a(2.6)
-81=-5.2a
a=81/5.2
a= 15.6m/s²
Angle of inclination =30°
To get the coefficient of friction, we use the formula
Ff =nR
Ff is frictional force
n is coefficient of friction
R is normal reaction
n = Ff/R = Wsin30°/Wcos30°
n = tan30°
n = 0.58
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Answer:
w = 1.976 rpm
Explanation:
For simulate the gravity we will use the centripetal aceleration , so:
where w is the angular aceleration and r the radius.
We know by the question that:
r = 60.5m
= 2.6m/s2
So, Replacing the data, and solving for w, we get:
W = 0.207 rad/s
Finally we change the angular velocity from rad/s to rpm as:
W = 0.207 rad/s = 0.207*60/(2)= 1.976 rpm