# How much work would it take to horizontally accelerate an object with a mass of 8 kg to 3 m/s on a surface with a kinetic friction coefficient of 3 ?

Jun 23, 2018

The work is $= 36 J$

#### Explanation:

The mass is $m = 8 k g$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

The coefficient of kinetic friction is ${\mu}_{k} = 3$

The normal force is $N = m g$

The force of friction is

${F}_{r} = {\mu}_{k} N = {\mu}_{k} m g$

According to Newton's Second Law

$F = m a$

The acceleration is

$a = \frac{F}{m} = \frac{{\mu}_{k} m g}{m} = {\mu}_{k} g$

The speed is $v = 3 m {s}^{-} 1$

The initial speed is $u = 0 m {s}^{-} 1$

Apply the equation of motion

${v}^{2} = {u}^{2} + 2 a s$

The distance is

$s = {v}^{2} / \left(2 a\right) = \frac{9}{2 \cdot {\mu}_{k} g} = \frac{9}{2 \cdot 3 \cdot 9.8} = 0.15 m$

The work done is

$W = F \cdot s = {\mu}_{k} m g \cdot s = 3 \cdot 8 \cdot 9.8 \cdot 0.15 = 36 J$