How much work would it take to horizontally accelerate an object with a mass of #4 kg# to #2 m/s# on a surface with a kinetic friction coefficient of #3 #?

1 Answer
Jul 19, 2017

Answer:

#W = 8# #"J"#

Explanation:

We're asked to find the work necessary to accelerate a #4#-#"kg"# object horizontally from rest to #2# #"m/s"# on a surface where the coefficient of kinetic friction is #3#.

To do this, we'll use the equation

#f_k = mu_kn#

where

  • #f_k# is the magnitude of the retarding friction force

  • #mu_k# is the coefficient of kinetic friction (#3#)

  • #n# is the magnitude of the normal force, which since the plane is horizontal is equal to

#n = mg = (4color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 39.24# #"N"#

The kinetic friction force is thus

#f_k = (3)(39.24color(white)(l)"N") = 118# #"N"#

This means a force of at least #118# #"N"# is needed to keep the object accelerating. The force doesn't matter, because the larger the force, the shorter the displacement once the object reaches #2# #"m/s"#.

The work will be the same regardless of the force here; let's say the applied force is #119# #"N"#, so the net force is

#119# #"N"# #-118# #"N"# #= 1# #"N"#

The magnitude of the acceleration is

#a_x = (sumF_x)/m = (1color(white)(l)"N")/(4color(white)(l)"kg") = 0.25# #"m/s"^2#

Using kinematics, we can find the displacement #Deltax# after it reaches #2# #"m/s"#, using the equation

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

It starts from rest, so we have

#(2color(white)(l)"m/s")^2 = 0 + 2(0.25color(white)(l)"m/s"^2)(Deltax)#

#Deltax = 8# #"m"#

Using

#W = Fs#

we have

#W = (1color(white)(l)"N")(8color(white)(l)"m") = color(red)(8# #color(red)("N"·"m"# #= color(red)(8# #color(red)("J"#