Explanation:
Given that,
Mass of the block, m = 12.2 kg
Initial velocity of the block, u = 6.65 m/s
The coefficient of kinetic friction, [tex]\mu_k=0.253[/tex]
(a)The force of kinetic friction is given by :
[tex]f=\mu_k mg[/tex]
mg is the normal force
So,
[tex]f=0.253\times 12.2\times 9.8\\\\f=30.24\ N[/tex]
(b) Net force acting on the block in the horizontal direction,
f = ma
a is the acceleration of the block
[tex]a=\dfrac{f}{m}\\\\a=\dfrac{30.24}{12.2}\\\\a=2.47\ m/s^2[/tex]
(c) Let d is the distance covered by the block before coming to the rest. Using third equation of motion as follows :
[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{-(6.65)^2}{2\times 2.47}\\\\d=-8.95\ m[/tex]
Hence, this is the required solution.
This question involves the concepts of the frictional force, Newton's second law of motion, and the equations of motion.
a) The force of kinetic friction acting on the block is "-30.28 N".
b) The acceleration of the block is "-2.48 m/s²".
c) The block will slide "8.9 m" before coming to rest.
a)
The kinetic friction force is given by the following formula:
[tex]F=\mu_k mg[/tex]
where,
F = frictional force = ?
[tex]\mu_k[/tex] = coefficient of kinetic friction = 0.253
m = mass of block = 12.2 kg
g = acceleration dueto gravity = 9.81 m/s²
Therefore,
[tex]F=(0.253)(12.2\ kg)(9.81\ m/s^2)[/tex]
F = - 30.28 N (negative sign due to resistive force)
b)
According to Newton's Second Law of Motion:
F = ma
[tex]a=\frac{F}{m} = \frac{-30.28\ N}{12.2\ kg}\\\\[/tex]
a = -2.48 m/s² (negative sign due to deceleration)
c)
Using the third equation of motion to calculate the distance:
[tex]2as=v_f^2-v_i^2[/tex]
where,
s = distance = ?
vf = final speed = 0 m/s
vi = initial speed = 6.65 m/s
Therefore,
[tex]2(-2.48\ m/s^2)s=(0\ m/s)^2-(6.65\ m/s)^2\\\\s=\frac{-(6.65\ m/s)^2}{2(-2.48\ m/s^2)}[/tex]
s = 8.9 m
Learn more about Newton's Second Law of Motion here:
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