# How much work would it take to horizontally accelerate an object with a mass of 4 kg to 3 m/s on a surface with a kinetic friction coefficient of 3 ?

Jan 10, 2018

$18 J$

#### Explanation:

In this case, kinetic friction is an opposing force directed opposite to the direction of motion.

For the above statement to be true, the acceleration acting on the object should be opposite (negative).

Forming an equation for frictional force,
$N {\mu}_{k} = M g \cdot {\mu}_{k}$
$N {\mu}_{k} = 40 \cdot 3 = 120$
where, $N$ = normal reaction = $M g$
$M$ = mass of the object

$\therefore$ frictional force = $120$

According to my second statement,
$120 = - M a$
$120 = - 4 a$

So, $a = - 30$m/${s}^{2}$

Now, substituting the above values in the equation,
${v}^{2} - {u}^{2} = 2 a s$

for an object to come at rest, its final velocity(v) must be = 0
So,

$0 - 9 = 2 \cdot - 30 \cdot s$

$s = \frac{3}{20} m$

Now,
$W = \vec{F} . \vec{s} \implies W = F s \cos \theta$

As applied force and displacement of object is in opposite direction,
$\theta = {180}^{\circ}$

$\therefore W = - M a \cdot s \cdot \cos {180}^{\circ}$

$W = 4 \cdot - 30 \cdot \frac{3}{20} \cdot - 1$

$W = 18 J$

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Hope This Helps