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Acceleration on an incline

  • Thread starter roam
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  • #1
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Homework Statement



[PLAIN]http://img237.imageshack.us/img237/6677/incline.gif [Broken]

A snow sled with a child secured safely to it has a total mass of 83.0 kg. It is lowered at a constant speed of 1.9 ms−1 down a slope of angle 41.0° with respect to the horizontal (as shown above) for a distance d = 14.0 m. The coefficient of kinetic friction between the sled and the snow is 0.09.

g = 9.8 ms–2. Air resistance is negligible at these speeds.

I have calculated the friction force to be 55.3, and the magnitude of the reactive force, N, on the sled to be 614.49 Newtons.

Then the question asks:

Suppose the rope suddenly snaps at its point of connection with the sled after travelling the distance d. Determine the drag force acting on the sled when it reaches a terminal speed of 8.4 ms−1.

The Attempt at a Solution



What does the question mean by the "drag force acting on the sled "? Is it reffering to the force that pulls the sled down the incline? I know that force is equal to

[tex]mg sin \theta = (83 \times 9.81) sin(41)= 534.18[/tex]

But this is wrong because the correct answer has to be 478 N.

Another approach I can think of is to find the time using dv=t= 117.6 and acceleration by [tex]a=v/t \Rightarrow a =\frac{8.4}{117.6}=0.07[/tex] then plug this into [tex]\sum F_x=ma_x=83 \times 0.07=5.9[/tex]. Which is again wrong... :blushing:

Can anyone help?
 
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Answers and Replies

  • #2
648
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At the terminal velocity, the component of the weight down the slope which you calculate to be 534.18N must be exactly balanced by the two frictional forces, the sliding friction, which you have calculated, and the air resistance; referred to as "drag".
 

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