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 #?

Question

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    #?

Elena Albarran · Accepted Answer

#W = 8# #"J"#

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"#

Layla Ahmed · Answer

undefined The work done to horizontally accelerate the object can be calculated using the formula:

$$ W = \frac{1}{2} m v_f^2 $$

where:
- $ W $ is the work done,
- $ m $ is the mass of the object (4 kg),
- $ v_f $ is the final velocity (2 m/s).

First, calculate the initial velocity using the friction force:

$$ F_f = \mu_k \cdot N $$

$$ a = \frac{F_f}{m} $$

$$ v_i = \sqrt{2a \cdot d} $$

$$ W = \frac{1}{2} m (v_f^2 - v_i^2) $$

where:
- $ F_f $ is the frictional force,
- $ \mu_k $ is the coefficient of kinetic friction (3),
- $ N $ is the normal force,
- $ a $ is the acceleration,
- $ v_i $ is the initial velocity,
- $ d $ is the distance over which the frictional force acts.

Given that the object starts from rest and there is no mention of the distance over which the frictional force acts, the work done cannot be accurately determined without that information.