What is the Work Done in a Ski Rescue?

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The discussion focuses on calculating the work done during a ski rescue involving a sled and victim on a slope. Key calculations include determining the work done by friction, the work done by the rope, and the gravitational work as the sled descends 30 meters at a 60-degree angle. Participants emphasize the importance of understanding the forces acting on the sled, including friction, and the need for a free body diagram to visualize these forces. The coefficient of friction is noted as 0.100, and the total mass of the sled and victim is 90 kg. Proper application of mechanical work definitions and relevant formulas is crucial for solving the problem accurately.
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Suppose the ski patrol lowers a rescue sled and victim 30m down a slope at a constant speed as shown below. The victim and sled have a total mass of 90kg. The angle of the slope is 60 degrees.

If the coefficient of friction is 0.100.
a) Find the work done by friction as the sled moves down the hill.
b) how much work is done by the rope on the sled.
c) What is the work done by gravitation.
d) What is the total work done.

Not sure where to start here. Any help with starting would be apprecitated.
 
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Make a schematic drawing with the sled and the slope. Draw all forces. Maybe it can help if you use two different colors for the forces exerted on the sled and the forces exerted by the sled. Compute the normal force and the friction force. Compute work done by the sled on the snow and on the rope. Do not forget, in the definition of mechanical work, the directions of forces and displacements.
 
Start with the first one by actually finding friction. And if you remember the definition of work, you should know to multiply it in the distance moved in that direction.

Same concept applies for the other 2 thought the final one is a bit different,
 
Formulas

Are there any formulas i should know for this particular problem. The thing that confuses me is that I know how to figure friction work, but i do not know how to find this with an angle involved.
 
The force due to friction is the coefficient of friction times the force perpendicular (normal) to the surface between the two things. If you are on the sled, the force to glide the sled over a horizontal snow will be 0.1 times your weight. But in the problem the surface is not horizontal. You must find the component of the weight perpendicular to the slope.
 
Constructing a proper free body diagram can be extremely useful in solving these types of problems. You can get some hints if you click on the first tutorial in my footer and open the pdf called motion3b.pdf
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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