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:
The change in momentum is 28265.71 kg-m/s.
Explanation:
Given that,
Mass of a car, m = 877 kg
Initial velocity of the car, u = 0 (at rest)
Final velocity of the car, v = 116 km/h = 32.23 m/s
Time, t = 0.951 s
We need to find the change in momentum produced by the force. It can be calculated as the difference of final momentum and the initial momentum.
So, the change in momentum is 28265.71 kg-m/s.