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

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" ")|)