7.8 | Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a
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- Опубліковано 12 вер 2024
- Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a 60.0º slope at constant speed, as shown in Figure 7.36. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by friction as the sled moves 30.0 m along the hill? (b) How much work is done by the rope on the sled in this distance? (c) What is the work done by the gravitational force on the sled? (d) What is the total work done?
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I actually found this problem to be pretty easy because it's primarily just the same concepts from forces on an inclined plane. Only here, we use those forces to find work. Goes to show how this stuff isn't too bad because it all builds on the material from earlier chapters. Very modular in nature and I like it that way. Good video as always man keep it up.
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In part A why did you remove the negative sign from Wy? When you input it back into the work formula you didn't carry the sign, why is that
Are you referring to how did the equation go from cos(6) = -wy/w to wy = -wcos(60)? This was just using algebra to solve for wy as the variable. To get wy by itself, we cross-multiplied and then just to pull the negative over to the other side. Hopefully this helps!
@@GlaserTutoring I was confused why you left the negative off, but after rewatching you specifically mentioned "equal in magnitude, but opposite in sign" to the normal force which clears up my confusion, thank you!
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Hello Sir. on formula for part a, why is not the displacement listed as negative since it is moving in negative X direction and F listed as positive since it is positive X direction?
Good effort
Faizan Ul Haq Thanks!