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I didn't go to the lesson of work-energy theorem, so I miss something about this subject. I know the formulas, but I can't figure it out. This question has many quantities.

Here is the problem,

The sled ($m = 11.1\;\mathrm{kg}$) shown in the figure leaves the starting point with a velocity of $25.1\;\mathrm{m/s}$. Use the work-energy theorem to calculate the sled’s speed at the end of the track or the maximum height it reaches if it stops before reaching the end. The straight sections of the track (A, B, D, and E) have a coefficient of friction of $0.409$ with the sled, and $284.9\;\mathrm{J}$ are lost to friction in the circular section of the track (C).

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Actually, In the inclined plane segment A, there is height and plane width, I should use

$$K_2 - K_1 = W$$

For A: $${m \over 2} v_\mathrm{f}^2 - {m \over 2} v_\mathrm{i}^2 = m g\cos(50^\circ) u_k d$$

$$v_\mathrm{f}=28.84\;\mathrm{m/s}$$

For B and D I used same equation without cos(degree)

For C part of it I used only change of kinetic energy

$$K_2 - K_1 = 284.9\;\mathrm{J}$$

My result is $26.46\;\mathrm{m/s}$

Answer is $24.112\;\mathrm{m/s}$

but I can't figure out the forces and quantities

By the way: I don't want the calculations I just want to understand the basic logic of it.

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3 Answers 3

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The sled ($m = 11.1\;\mathrm{kg}$) shown in the figure leaves the starting point with a velocity of $25.1\;\mathrm{m/s}$. Use the work-energy theorem to calculate the sled’s speed at the end of the track or the maximum height it reaches if it stops before reaching the end. The straight sections of the track (A, B, D, and E) have a coefficient of friction of $0.409$ with the sled, and $284.9\;\mathrm{J}$ are lost to friction in the circular section of the track (C).

It is very simple :

  • the sled leaves with $KE_i = 0.5*11.1*25.1^2$ at $h=30m$,

  • at the end of the path it is at $h_f= 10m$. Supposing $g=10$ the energy gained by the drop is $KE_h= mg(h-h_1) = 111*(30-10)$, the sum of $KE_i+KE_h$ would be the energy of the sled in absence of friction, but we must subtract $W_C=284.9 J$ lost in C plus

  • the energy lost to friction in B+D = 15 $W_{BD}= \mu mg*d= 0.409*111*15$
  • the energy lost to friction in A+E = 52.22m $W_{AE}=\mu mg*d*cos50= 0,409*111*52.22*0.64$

The difference is the KE of the sled at the end of the track and you find that its speed is 24.112

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Try to keep this tidy. It is a straight forward calculation, but there are many terms, so tidiness is the key. Start with the Energy work relation:

$$E_A - E_E = W$$

where $E_A$ is the energy at the beginning, $E_E$ the energy at the end and $W$ the energy loss due to friction. We have to split $W$ further into

$$W = W_A + W_B + W_C + W_D + W_E$$

where $W_X$ denotes the energy loss for the part $X$.

What you already have is $W_A$:

$$W_A = \mu \cdot m \cdot g \cdot \frac{H}{\sin(\alpha)} \cdot \cos(\alpha)$$

I call $\mu$ the friction coefficient, the height $H$ and $\alpha$ is the only angle given in the drawing.

$\frac{H}{\sin(\alpha)} = d $ in your naming convention. I prefer to keep as few variables as possible.

What are the other quantities? $W_B$ and $W_D$ are easy, $W_C$ is already given, $W_E$ is calculated the same way as $W_A$, hence we are almost there. $E_A$ and $E_E$ are given as sum of kinetic and potential energies. A bit of algebra, and you have the result.

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I am not so sure about your first step. One sees a gain in speed due to conversion of potential energy into kinetic energy. That would give

${{m}\over{2}} v^2_\mathrm{f}-{{m}\over{2}} v^2_\mathrm{i}= m g h$

no need for angular functions as the hight is already shown in sketch, but you must remove losses due to friction as well. Looking at the standard definition of friction forces, you get $F_\mathrm{friction}$ and the energy loss is the integral of the force along the path, in this case just $E_\mathrm{loss\; on\; A}=FA$. So you get

${{m}\over{2}} v^2_\mathrm{f}-{{m}\over{2}} v^2_\mathrm{i}= m g h-E_\mathrm{loss\; on\; A}$

I leave it to you to calculate the friction force and the according energy loss.

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  • $\begingroup$ Energy loss on A could be $$ m.g.cos(50 degree).X.u_k $$ ? (X is the path on meter) $\endgroup$
    – alim
    Apr 23, 2015 at 17:51
  • $\begingroup$ @alim If $u_k$ is you kinetic friction coefficient, yes. $\endgroup$ Apr 24, 2015 at 7:22

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