Physics, 11.10.2020 14:01 xojade

A block of mass 12.2 kg is sliding at an initial velocity of 6.65 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.253. (Indicate the direction with the signs of your answers.) a.) What is the force of kinetic friction (in N) acting on the block?
(b What is the block's acceleration (in m/s2)?

(c) How far will it slide (in m) before coming to rest?

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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.

The kinetic friction force is given by the following formula:

$F=\mu_k mg$

where,

F = frictional force = ?

$\mu_k$ = coefficient of kinetic friction = 0.253

m = mass of block = 12.2 kg

g = acceleration dueto gravity = 9.81 m/s²

Therefore,

$F=(0.253)(12.2\ kg)(9.81\ m/s^2)$

F = - 30.28 N (negative sign due to resistive force)

According to Newton's Second Law of Motion:

F = ma

$a=\frac{F}{m} = \frac{-30.28\ N}{12.2\ kg}\\\\$

a = -2.48 m/s² (negative sign due to deceleration)

Using the third equation of motion to calculate the distance:

2as=v_f^2-v_i^2

where,

s = distance = ?

vf = final speed = 0 m/s

vi = initial speed = 6.65 m/s

Therefore,

$2(-2.48\ m/s^2)s=(0\ m/s)^2-(6.65\ m/s)^2\\\\s=\frac{-(6.65\ m/s)^2}{2(-2.48\ m/s^2)}$

s = 8.9 m

Learn more about Newton's Second Law of Motion here:

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The attached picture shows Newton's Second Law of Motion.

Answer from: pineapplepizaaaaa

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, $\mu_k=0.253$

(a)The force of kinetic friction is given by :

$f=\mu_k mg$

mg is the normal force

So,

$f=0.253\times 12.2\times 9.8\\\\f=30.24\ N$

(b) Net force acting on the block in the horizontal direction,

f = ma

a is the acceleration of the block

$a=\dfrac{f}{m}\\\\a=\dfrac{30.24}{12.2}\\\\a=2.47\ m/s^2$

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{-(6.65)^2}{2\times 2.47}\\\\d=-8.95\ m$

Hence, this is the required solution.

A block of mass 12.2 kg is sliding at an initial velocity of 6.65 m/s in the positive x-direction. T

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A block of mass 12.2 kg is sliding at an initial velocity of 6.65 m/s in the positive x-direction. T...