The Drag Force at Terminal Velocity

In summary, a snow sled with a child secured safely to it has a total mass of 83.0 kg and is lowered down a slope of angle 41.0° at a constant speed of 1.9 ms−1 for a distance of 14.0 m. The coefficient of kinetic friction between the sled and the snow is 0.09, and air resistance is negligible. After the sled travels the distance d, the rope snaps and the sled reaches a terminal speed of 8.4 ms−1. The question asks for the drag force acting on the sled at this point, which must balance out the weight and frictional forces.
  • #1
roam
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Homework Statement



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

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 traveling 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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  • #2
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".
 

1. What is acceleration on an incline?

Acceleration on an incline refers to the rate at which an object's velocity changes when it is moving on a surface that is not horizontal. It is affected by the angle of the incline, the object's mass, and the force of gravity.

2. How is acceleration calculated on an incline?

The formula for calculating acceleration on an incline is a = g*sin(theta), where "a" is the acceleration, "g" is the acceleration due to gravity (9.8 m/s^2), and "theta" is the angle of the incline.

3. Does acceleration on an incline affect an object's speed?

Yes, acceleration on an incline affects an object's speed because it determines the rate at which the object's velocity changes. The steeper the incline, the greater the acceleration, and the faster the object will gain speed.

4. How does the mass of an object affect acceleration on an incline?

The mass of an object does not directly affect acceleration on an incline. However, a heavier object may require more force to move up the incline, which can impact the object's acceleration.

5. Can an object have negative acceleration on an incline?

Yes, an object can have negative acceleration on an incline if it is slowing down while moving up the incline. This is because acceleration is a vector quantity and can have a negative direction, depending on the object's motion.

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