Question

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?

Answer 1

`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)`

Answer 2

`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.