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3 years ago

15

A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is

0.15 and the push imparts an initial speed of 3.5m/s?

2 answers:

ollegr [7]3 years ago

8 0

<span>F = ma
</span>Ff = μ*Fn
<span>Fn = Fw
</span>Fw = mg

<span>So we have: </span>

<span>Ff = μmg </span>

<span>And </span>

<span>Ff = ma </span>

<span>So... </span>

<span>μmg = ma </span><span> </span>

<span>μg = a </span>

<span>And we can solve for the acceleration: </span>

<span>(0.15)(9.81 m/s²) = a </span>

<span>a = 1.47 m/s² </span>

Gre4nikov [31]3 years ago

7 0

Answer:

The answer is:

xf = 4.17m

Explanation:

Hello!

Let's solve this!

First we have to calculate the acceleration, then calculate the distance the box reaches.

The force is:

F = m * a

Frictional force (ff)

ff = μ * Fn

Fn = Fw

Fw = mg

Then we match

ff = μ * m * g

ff = m * a

So:

μmg = ma

μg = a

(0.15) (9.81 m / s²) = a

a = 1.47 m / s²

Then we know that the formula is:

vf2 = vi2 + 2a (xf-xi)

0 = (3.5m / s) 2 + 2 * 1.47 m / s² * (xf-0)

xf = - (3.5m / s) 2 / (- 2 * 1.47 m / s²)

The answer is:

xf = 4.17m

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