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 #6 #?

1 Answer
Aug 18, 2017

Answer:

#W = 36# #"J"# #+ 471color(white)(l)"N" xx "distance"#

Explanation:

We're asked to find the necessary work to accelerate an #8#-#"kg"# object from #0# to #3# #"m/s"# on a surface with a coefficient of kinetic friction of #6#.

The equation for work according to the work-energy theorem is given by

#ul(W = DeltaE_k = 1/2m(v_2)^2 - 1/2m(v_1)^2#

The final velocity is #3# #"m/s"#, and its mass is #8# #"kg"#, so we have

#W = 1/2(8color(white)(l)"kg")(3color(white)(l)"m/s" - 0)^2 = color(red)(ul(36color(white)(l)"J"#

However, this does not include the work done to counteract the friction, so we have to use the equation

#ul(W = Fs#

The necessary force #F# is

#F = f_k = mu_kmg = (6)(8color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 471# #"N"#

And so we have

#W = (471color(white)(l)"N")s#

The distance it travels #s# can be any value because the applied force can be any number greater than the necessary force. Thus, we can leave it in terms of #s#:

#W_"necessary" = color(blue)(ulbar(|stackrel(" ")(" "36color(white)(l)"J" + (471color(white)(l)"N")s" ")|)#