Question

A box with an initial speed of `1 m/s` is moving up a ramp. The ramp has a kinetic friction coefficient of `2/3` and an incline of `(5 pi )/12`. How far along the ramp will the box go?

Answer 1

About `4.5"cm"`.

Explanation:

There are two main methods we can use to solve this: a combination of Newton's second law and kinematics, or the work-energy theorem. I will show both—this will be a lengthly answer.

Second Law and Kinematics

Here we can begin with a basic force diagram.

Let's define up the ramp as the negative direction. Then we can set up statements for the net force parallel and perpendicular.

• `sumF_"//"=f_k+F_(g"//")=ma_"//"`

• `sumF_"⟂"=n-F_(g"⟂")=ma_"⟂"`

Since the box is slowing down, we know that it is experiencing an acceleration in the opposite direction of its motion, or in this case down the ramp. This is entirely parallel, or in the x-direction.

Since the box does not move vertically relative to the ramp, we know that `a_"⟂"=0`. Hence, we have:

• `f_k+F_(g"//")=ma_"//"`

• `n-F_(g"⟂")=0`

• `=>n=F_(g"⟂")`

We know that the force of kinetic friction `f_k` is given by `f_k=mu_kn`, and the force of gravity by `F_g=mg`.

Since we are on an incline, we have to break the force of gravity up into its parallel and perpendicular components, as shown in the diagram. We can do this using trigonometry.

• By the magic of geometry, the angle between the gravitational force and the vertical is the same as the angle of incline.

• Therefore, we can see that the parallel component of the gravitational force is opposite the angle, the perpendicular component is adjacent to the angle, and the overall magnitude is the hypotenuse.

Consequently, we have:

• `F_(g"//")=F_gsin(theta)=mgsin(theta)`

• `F_(g"⟂")=F_gcos(theta)=mgcos(theta)`

Revisiting our sum of forces statements:

• `mu_kn+mgsin(theta)=ma_"//"`

• `n=mgcos(theta)`

• `=>mu_kmgcos(theta)+mgsin(theta)=ma_"//"`

We can then solve for the acceleration parallel:

`a=(mu_kcancelmgcos(theta)+cancelmgsin(theta))/cancelm`

`=>color(blue)(a=g(sin(theta)+mu_kcos(theta)))`

We can then use a kinematic equation to derive an expression for the distance traveled up the ramp:

• `v_f^2=v_i^2+2aDeltas`

Solving for `Deltas`:

`=>Deltas=(-v_i^2)/(2a)`

Substituting in our expression for acceleration found above:

`=>color(blue)(Deltas=(-v_i^2)/(2g(sin(theta)+mu_kcos(theta))))`

Substituting in our known values:

`Deltas=(-(1"m"/"s")^2)/(2(9.81"m"/"s"^2)(sin((5pi)/12)+(2/3)cos((5pi)/12))`

`=>=-0.045m`

Note this quantity is negative because we have defined up the ramp as the negative direction.

`=>color(blue)(Deltas=4.5"cm")` up the ramp.

Work-Energy Theorem

Our basic model of energy conservation is as follows:

`E_"system"=E_"mechanical"+E_"thermal"=K+U+E_"th"`

• When energy is conserved, we have `DeltaE_"sys"=0`.

• When energy is not conserved, `DeltaE_"sys"!=0`, which implies that we have energy transfer. The mechanical transfer of energy to or from the system is called work.

The work-energy theorem states that, in summary, the change in mechanical energy of a system is due to work being done by nonconservative forces—the most popular of which is friction.

`color(blue)(DeltaK+DeltaU=DeltaE_"mech"=W_"nc")`

Therefore, it must be in our case that `DeltaE_"mech"=W_"friction"`.

• Essentially, we say that if energy is not conserved, it must be because there is some outside force "robbing" energy from the system (or perhaps supplying it)

Work is given by the dot product of the force and displacement vectors, which for a constant force becomes:

`W=Fdcos(theta)`

Where `theta` is the angle between the force and displacement vectors.

• In this case, `theta=180^o` as the displacement of the box is up the ramp, whereas the force of friction is directed down the ramp. In other words, they are anti-parallel.

Since `cos(180^o)=-1`, we have:

`color(blue)(W_f=-f_k*d)`

Initially, the box is moving up the ramp, so it has initial kinetic energy. Because it comes to rest, it has no kinetic energy finally.

Similarly, the box begins at the bottom of the ramp, where we define `h=0`, so there is no gravitational potential energy. However, when the box comes to rest at some distance up the ramp, it has some final height and hence gravitational potential energy.

• `DeltaK=K_f-K_i=0-1/2mv_i^2`

• `DeltaU=U_f-U_i=mgh_f-0`

Then we have:

`-1/2mv_i^2+mgh_f=-f_k*d`

We now note that the height up the ramp `h_f` can be expressed in terms of the distance up the ramp using trigonometry.

• `h=dsin(theta)`

`=>-1/2mv_i^2+mgdsin(theta)=-f_k*d`

We can solve for `d`:

`1/2mv_i^2=f_kd+mgdsin(theta)`

`=>1/2cancelmv_i^2=d(mu_kcancelmgcos(theta)+cancelmgsin(theta))`

`=>color(blue)(d=(v_i^2)/(2g(mu_kcos(theta)+sin(theta))))`

Which is the same expression that we derived above using kinematics and Newton's second.

`=>color(blue)(d=4.5"cm")` up the ramp.