A factory worker accidentally releases a 180 kgcrate that was being held at rest at the top of a ramp that is 3.7 m long and inclined at $\mathbf{39}\mathbf{°}$to the horizontal. The coefficient of kinetic friction between the crate and the horizontal factory floor is 0.28. (a) How fast is the crate moving as it reaches the bottom of the ramp? (b) How far will it subsequently slide across the floor? (Assume that the crate’s kinetic energy does not change as it moves from the ramp onto the floor.) (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we halve the mass of the crate?

The mass of the crate is,m=180 kg

The length of a ramp is, L=3.7 m

The angle of inclination is,$\theta =39°$

The coefficient of friction,$\mathrm{\mu }=0.28$

The initial velocity is,$v_{\mathrm{i}}=0\mathrm{m}/\mathrm{s}$

Using the formula for workdone on a system by external force when it is on the ramp, we can find the velocityof the crate at the bottom of the ramp. Then, using the formula for workdone on a system by external force when it is sliding along the floor and usingtheresult obtained from part a) we can findthe distance through which the crate slides across the floor.

Formulae:

The work done by the body,$W=\Delta E_{mech}+\Delta E_{th}$ (1)

The thermal energy due to friction,$\Delta E_{th}={\mu }_{kNL}$ (2)

The potential energy at a height,$PE=mgh$ (3)

The kinetic energy of the body,$\mathrm{KE}=\frac{1}{2}m{v}^2$ (4)

In this case,W=0 Then, the thermal energy due to friction is given using equation (2):

$\Delta E_{th}={\mu }_{k}mgcos\theta L...........\left(5\right)$

$(N=mg\cos \theta ,fortheramp)$

From trigonometry we can write for the given system that, the change in potential energy is given as:

$\Delta P.E.=-mgL\sin \theta ...................\left(6\right)$

Thus, using equations (5) and (6) in equation (1), the final velocity of the crate is given as:

$$\begin{gathered} 0\hspace{0.33em}J=\Delta K.E.+\Delta P.E.+\Delta E_{th} \\ 0\hspace{0.33em}J=K.E{.}_f-K.E{.}_i+-mgLsin\theta +{\mu }_{k}mgcos\theta L \\ 0\hspace{0.33em}J=K.E_f-0+-mgLsin\theta +{\mu }_{k}mgcos\theta L \\ K.E{.}_f=mgLsin\theta -{\mu }_{k}mgcos\theta L \\ \frac{1}{2}m{v}_f^2=mgLsin\theta -{\mu }_{k}mgcos\theta L \end{gathered}$$

$$\begin{gathered} {\mathrm{V}}_{\mathrm{f}}=\sqrt{\frac{2}{\mathrm{m}}\left(\mathrm{mgLsin\theta }-{\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta L}\right)} \\ =\sqrt{2\mathrm{gL}\left(\mathrm{sin\theta }-{\mathrm{\mu }}_{\mathrm{k}}\mathrm{cos\theta }\right)} \\ =\sqrt{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(3.7\mathrm{m}\right)\left(\sin 39°-0.28\cos 39°\right)} \\ =5.46\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the velocity of the crate at the bottom of the ramp is $5.46\frac{m}{s}$.

When the crate moves along the floor a distance$L\text{'}$thework done on a system by the external force is zero. Thus, the distance traveled by the crate using equation (1) is given as:

$$\begin{gathered} 0\hspace{0.33em}J=\Delta E_{mech}+\Delta E_{th} \\ 0\hspace{0.33em}J=K.E{.}_f-K.E{.}_i+0\hspace{0.33em}J+{\mu }_{k}mg{L}^{\text{'}} \end{gathered}$$

The velocity obtained in part (a) is the initial velocity of the crate on the factory floor. Thus, substituting the value we get the distance as:

$$\begin{gathered} 0\hspace{0.33em}J=0\hspace{0.33em}J-\frac{1}{2}m{v}_f^2+{\mu }_{k}mg{L}^{\text{'}} \\ L\text{'}=\frac{{\displaystyle \frac{1}{2}}{v}_f^2}{{\mu }_{k}g} \\ L\text{'}=\frac{{\displaystyle \frac{1}{2}}{\left(5.46{\displaystyle \frac{m}{s}}\right)}^2}{\left(0.28\right)\left(9.8{\displaystyle \frac{m^2}{s}}\right)} \\ L\text{'}=5.43m \end{gathered}$$

Therefore, the distance through which it slides across the floor is 5.43 m

From equations (1) and (2) obtained in parts (a) and (b) we can conclude that answers for parts (a) and (b) do not depend on the mass of the crate.

Therefore, the answers for parts (a) and (b) remain the same when the mass of the crate is halved.