If a #3/2 kg# object moving at #5/3 m/s# slows to a halt after moving #4/3 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

2 Answers
Aug 20, 2017

Answer:

#mu_k=0.11#

Explanation:

We are given the following information:

  • #|->m=3//2" kg"#
  • #|->Deltas=4//3" m"#
  • #|->v_i=5//3" m"//"s"#
  • #|->v_f=0#
  • #|->g=9.81" m"//"s"^2#

#=>#This problem can be solved using either kinematics and Newton's second law, or the work-energy theorem. I will include both solutions.

**#color(darkblue)("Method 1: Newton's Second Law and Kinematics.")#

  • We can use the following kinematic equation to solve for the acceleration that the object experiences as it comes to rest:

#color(blue)(v_f^2=v_i^2+2aDeltas)#

  • After calculating the acceleration, we can generate a statement of the net force on the object using Newton's second law:

#color(blue)(vecF_"net"=mveca)#

  • And we can use the acceleration and given mass to calculate the net force: the force of kinetic friction #f_k#.

Let's solve for #a# in the kinematic:

#=>color(blue)(a=(v_f^2-v_i^2)/(2Deltas))#

Using our known values:

#=>a=(-(5/3" m"//"s")^2)/(2*4/3"m")#

#=>color(blue)(a~~-1.042" m"//"s"^2)#

Note that the sign of the acceleration is negative, which indicates that the acceleration is the opposite direction of motion and therefore the object is slowing down.

  • As for the net force, the perpendicular (y, vertical) forces include only the normal force and force of gravity, which are in a state of equilibrium. The parallel (x, horizontal) forces, however, include only the force of kinetic friction, which is what causes the object to slow down.

#F_(x" net")=-f_k=ma#

The statement of the perpendicular forces in equilibrium #(a=0)# is:

#F_(y" net")=n-F_G=0#

As we know that #F_G=mg:#

#=>n=mg#

We also know that #f_k=mu_kn#

#=>f_k=mu_kmg#

Therefore:

#-mu_kmg=ma#

  • Solving for #mu_k#:

#=>-mu_k=(ma)/(mg)#

#=>-mu_k=a/g#

  • Using our known values:

#-mu_k=(-1.042" m"//"s"^2)/(9.81"m"//"s"^2)#

#=>color(blue)(0.11)#

**#color(darkblue)("Method 2: Work-Energy Theorem.")#

By the work-energy theorem, the work done by nonconservative forces (e.g. the force of friction) is equal to the energy lost in a system.

#color(blue)(DeltaE_"sys"=W_"nc")#

When energy is conserved in a system, #DeltaE_"sys"=0#.

We have only kinetic energy, so this statement becomes:

#W_"friction"=DeltaK#

#=>W_F=1/2mv_f^2-1/2mv_i^2#

There is no kinetic energy finally, as the object is at rest and #v=0#:

#=>W_F=-1/2mv_i^2#

The work done by friction:

#W_F=f_kDeltascos(theta)#

  • Where #f_k# is the force of kinetic friction, #Deltas# is the displacement of the object, and #theta# is the angle between the force and displacement vectors. In this case, friction is opposite the motion and therefore antiparallel, so #cos(180^o)=-1#

#=>W_F=-f_kDeltas#

#=>-1/2mv_i^2=-mu_knDeltas#

#=>1/2mv_i^2=mu_kmgDeltas#

We can now solve for #mu_k#:

#=>mu_k=(1/2mv_i^2)/(mgDeltas)#

#=>color(blue)(mu_k=(1/2v_i^2)/(gDeltas))#

Using our known values:

#=>mu_k=(1/2(5/3"m"/"s")^2)/((9.81"m"//"s"^2)(4/3"m"))#

#=>mu_k=0.106#

#=>color(blue)(mu_k~~0.11)#

Aug 20, 2017

Answer:

#mu = 0.107#

Explanation:

We can find the acceleration of the object using the kinematic formula
#v^2 = u^2 + 2*a*s#
#0^2 = (5/3 m/s)^2 + 2*a*4/3 m #
Solving that for a,
#-2*a*4/3 m = (5/3 m/s)^2#
#a = -(5^2 * 3)/(2 * 3^2 * 4) m^2/(m*s^2)#
#a = -(5^2 * cancel(3))/(2 * 3^cancel(2) * 4) m^cancel(2)/(cancel(m)*s^2) = -25/24 m/s^2#

The force responsible for that acceleration must have been
#F = m*a = 3/2 kg*(-25/24 m/s^2)#
#F = -75/48 N#

(I will assume that the surface this thing is on is horizontal since there is no information to tell me that I have to make it more complicated.)
The friction formula is
#F_f = mu*N = mu*m*g#
#-75/48 N = mu*3/2 kg*9.8 m/s^2#
Solving that for #mu#,
#mu = -(75/48 N) / (3/2 kg*9.8 m/s^2) = -(75/48 N) / ((3/2)*9.8 N)#
#mu = 0.107#
Note, the negative sign indicates that the direction of #Ff# is in the opposite direction of the initial velocity. But the coefficient of kinetic friction is a scalar, so I have dropped the sign.

I hope this helps,
Steve
p.s. double-check my arithmetic.