A 1,800 kg car is parked on a road that has an elevation level of 7°. Suppose the coefficient of static friction is .65. What is
the force of static friction?
1 answer:
m = mass of the car = 1800 kg
g = acceleration due to gravity = 9.8 m/s²
= force of gravity on the car
force of gravity on the car is given as
= mg
inserting the values
= (1800) (9.8) = 17640 N
= static frictional force acting on the car
from the force diagram of the car , the static frictional force is opposite to the parallel component of force of gravity on the car . hence static frictional force mus balance the component of force of gravity parallel to incline surface.
hence
=
Sin7
= (17640) Sin7
= 2149.8 N
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Answer:
1.52905 seconds
4.58715 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
= Coefficient of friction = 0.4
g = Acceleration due to gravity = 9.81 m/s²
a = Acceleration =
Equation of motion
It will take 1.52905 seconds for the block to slow down
The block will travel 4.58715 m before it stops