# How much work would it take to horizontally accelerate an object with a mass of 6 kg to 7 m/s on a surface with a kinetic friction coefficient of 3 ?

Jun 26, 2017

The work is $= 147 J$

#### Explanation:

The coefficient of kinetic friction is

${\mu}_{k} = {F}_{r} / \left(N\right)$

The normal force is

$\left(N\right) = m g = 6 g N$

The frictional force is

${F}_{r} = {\mu}_{k} \cdot \left(N\right) = 3 \cdot 6 g = 18 g N$

According to Newton's Second Law, the acceleration is

$F = m a$

$a = \frac{F}{m} = 18 \frac{g}{6} = 3 g$

To find the distance, we apply the equation of motion

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

$u = 0$

$s = {v}^{2} / \left(2 a\right) = {7}^{2} / \left(2 \cdot 3 g\right) m$

The work is

$W = F s = 18 g \cdot \frac{49}{6 g} = 147 J$