91Ó°ÊÓ

The mass of the crate is,m=180 kg

The length of a ramp is, L=3.7 m

The angle of inclination is,$\mathrm{θ}=39°$

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

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,$W=\mathrm{Δ}{E}_{mech}+\mathrm{Δ}{E}_{th}$ (1)

The thermal energy due to friction,$\mathrm{Δ}{E}_{th}={\mathrm{μ}}_{kNL}$ (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):

$\mathrm{Δ}{E}_{th}={\mathrm{μ}}_{k}mgcos\mathrm{θ}L...........\left(5\right)$

$(N=mg\cos \mathrm{θ},fortheramp)$

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

$\mathrm{Δ}P.E.=-mgL\sin \mathrm{θ}...................\left(6\right)$

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

$$\begin{gathered} 0 J=\mathrm{Δ}K.E.+\mathrm{Δ}P.E.+\mathrm{Δ}{E}_{th} \\ 0 J=K.E{.}_f-K.E{.}_i+-mgLsin\mathrm{θ}+{\mathrm{μ}}_{k}mgcos\mathrm{θ}L \\ 0 J=K.E_f-0+-mgLsin\mathrm{θ}+{\mathrm{μ}}_{k}mgcos\mathrm{θ}L \\ K.E{.}_f=mgLsin\mathrm{θ}-{\mathrm{μ}}_{k}mgcos\mathrm{θ}L \\ \frac{1}{2}m{v}_f^2=mgLsin\mathrm{θ}-{\mathrm{μ}}_{k}mgcos\mathrm{θ}L \end{gathered}$$

$$\begin{gathered} {\mathrm{V}}_{\mathrm{f}}=\sqrt{\frac{2}{\mathrm{m}}\left(\mathrm{³¾²\mu ³¢²õ¾\pm ²Ôθ}-{\mathrm{μ}}_{\mathrm{k}}\mathrm{³¾²\mu ³¦´Ç²õθ³¢}\right)} \\ =\sqrt{2\mathrm{gL}\left(\mathrm{²õ¾\pm ²Ôθ}-{\mathrm{μ}}_{\mathrm{k}}\mathrm{³¦´Ç²õθ}\right)} \\ =\sqrt{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(3.7\mathrm{m}\right)\left(\sin 39°-0.28\cos 39°\right)} \\ =5.46\mathrm{m}/\mathrm{s} \end{gathered}$$

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:

$$\begin{gathered} 0 J=\mathrm{Δ}{E}_{mech}+\mathrm{Δ}{E}_{th} \\ 0 J=K.E{.}_f-K.E{.}_i+0 J+{\mathrm{μ}}_{k}mg{L}^{\text{'}} \end{gathered}$$

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:

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

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.