Question

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 `1`?

Answer 1

`147J`

Explanation:

Concept 1
In the problem, it is important to note that there is no friction. In the formula,

`F_f=muF_N`

`mu` represents the constant of proportionality. But since it equals to `1`, `F_f=F_N`.

Concept 2
To solve the problem, you must use two formulas:

`color(blue)(|bar(ul(color(white)(a/a)W=DeltaE_kcolor(white)(a/a)|)))`
where:
`W=`work (joules)
`DeltaE_k=`change in kinetic energy (joules)

`color(blue)(|bar(ul(color(white)(a/a)E_k=1/2mv^2color(white)(a/a)|)))`
where:
`m=`mass (metres)
`v=`velocity (metres per second)

Solving for Work
Start by breaking down `DeltaE_k` into final kinetic energy minus the initial kinetic energy.

`W=DeltaE_k`

`W=E_(k,"final")-E_(k,"initial")`

`W=1/2mv_f^2-1/2mv_i^2`

`W=1/2m(v_f-v_i)^2`

Substitute your values.

`W=1/2(6kg)(7m/s-0m/s)^2`

Solve.

`W=1/2(6kg)(49m^2/s^2)`

`W=color(green)(|bar(ul(color(white)(a/a)color(black)(147J)color(white)(a/a)|)))`