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 ?

1 Answer
May 14, 2016

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)|)))