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 #1 #?

1 Answer
May 14, 2016

Answer:

#36J#

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(8kg)(3m/s-0m/s)^2#

Solve.

#W=1/2(8kg)(3m/s)^2#

#W=color(green)(|bar(ul(color(white)(a/a)color(black)(36J)color(white)(a/a)|)))#