# 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

#### Answer:

#### 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#

**kinematics** and **Newton's second law**, or the **work-energy** theorem. I will include both solutions.

**

- 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

#=>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

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

As we know that

#=>n=mg#

We also know that

#=>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)#

**

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,

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

#=>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=(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)#

#### Answer:

#### Explanation:

We can find the acceleration of the object using the kinematic formula

Solving that for a,

The force responsible for that acceleration must have been

(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

Solving that for

Note, the negative sign indicates that the direction of

I hope this helps,

Steve

p.s. double-check my arithmetic.