How much work would it take to horizontally accelerate an object with a mass of #8 kg# to #7 m/s# on a surface with a kinetic friction coefficient of #9 #?

1 Answer
Jul 29, 2017

Answer:

#W = 196# #"J"#

Explanation:

We're asked to find the total work done on an object so that it accelerates to #7# #"m/s"#, with a coefficient of kinetic friction of #9#.

Using the equation

#f_k = mu_kn#,

we can find the retarding friction force the surface exerts on the object (knowing that the normal force #n = mg#):

#f_k = mu_kmg = (9)(8color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = color(red)(706.32# #color(red)("N"#

To make this simple, we'll make the net force experienced by the object #1# #"N"#, so the applied force will be #707.32# #"N"#:

#sumF_x = 707.32# #"N"# #- 706.32# #"N"# #= 1# #"N"#

Now, we can use Newton's second law to find the acceleration of the object:

#sumF_x = ma_x#

#a_x = (sumF_x)/m = (1color(white)(l)"N")/(8color(white)(l)"kg") = 0.125# #"m/s"^2#

Now, let's use a kinematics equation to find the distance traveled under this constant acceleration so that the final velocity is #7# #"m/s"#:

#(v_x)^2 = (v_(0x))^2 + 2a_x(Deltax)#

Starting from an assumed state of rest:

#(7color(white)(l)"m/s")^2 = 0 + 2(0.125color(white)(l)"m/s"^2)(Deltax)#

#Deltax = 196# #"m"#

Finally, using the equation

#W = Fs# (one dimensional)

(#F# is the net force), we have

#W = (1color(white)(l)"N")(196color(white)(l)"m") = color(blue)(ul(196color(white)(l)"J"#