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:
a= - 6.667 m/s²
Explanation:
Given that
The initial speed of the box ,u= 20 m/s
The final speed of the box ,v= 0 m/s
The distance cover by box ,s= 30 m
Lets take the acceleration of the box = a
We know that
v²= u ² + 2 a s
Now by putting the values in the above equation we get
0²=20² + 2 a x 30
a= - 6.667 m/s²
Negative sign indicates that velocity and acceleration are in opposite direction.
Therefore the acceleration of the box will be - 6.667 m/s² .