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$\bullet$$\bullet$ You are working for a shipping company. Your job is tostand at the bottom of an 8.0 -m-long ramp that is inclined at$37^{\circ}$ above the horizontal. You grab packages off a conveyorbelt and propel them up the ramp. The coefficient of kineticfriction between the packages and the ramp is $\mu_{k}=0.30$(a) What speed do you need to give a package at the bottom ofthe ramp so that it has zero speed at the top of the ramp?(b) Your coworker is supposed to grab the packages as theyarrive at the top of the ramp, but she misses one and it slidesback down. What is its speed when it returns to you?

a) 11.48 $\mathrm{m} / \mathrm{s}$b) 7.52 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 5

Applications of Newton's Law

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Simon Fraser University

University of Sheffield

University of Winnipeg

Lectures

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All right, So the free body diagrams for part A and party B are shown here on DH. You can see the little force and gravity. Obviously, there are the same. But what's different is Azad in part, air velocity is acting upwards is going up the the thing is moving upwards, and so you're frictional force will be acting downwards to resist that motion. In part, be official force will be acting upwards because your your boxes sliding down. So motion it stop so anywhere they're two steps too, to solving boat parts. And those are first find acceleration from Newton's second law and then, ah, find velocity from our initial velocity from an equation of motion. Okay. And so, uh, let's go ahead and and get started in. So we resolve forces with in the white direction and there's no net force in the white direction. So important air. You have n equals. The normal force equals the vertical component of gravity. So mg co sign data the normal force. Ah. Then you do that for the why and the wider action you have. Ah, sorry. The extraction. You have Ah, not acceleration in the X erection and So here you have. Um, you have mg cosa ngo signed data and that's a negative. MG signed data the ex component after after of the force of gravity minus the frictional force. That's a care because they double back downwards and downward is negative for us is equal to m times acceleration, the ex direction. And so this's just m Jeez, they haven't decided. Data minus of sub care is music a times and and we know n is mg co signed data s O. This is Ali. Quarter and turns is X And so as you can see the ends really, I'll drop out. So you left with a setbacks calls negative. Two times signed data plus music care Times co signed data. So this is negative. 9.8 time sign 37 plus music K is Mr K is 0.3. So 0.3 signed 37 and what you get is a minus eight point 25 minus 8.25 meters per second square. This means acceleration is, uh, acceleration is downwards. And this magnitude this 8.5 meters per second squared on DH. So for the next part, we use an equation of motion and this will be the squared equals B not squared, plus two beeswax tons X um and so we have X on. We have a sex. So we want you not squared b squared and finally comes. It's it goes to rest and so back goes away. And so we not squared is negative. Two times a Saxon times X and so that's negative. Out two times negative. 8.25 minutes. We just found times X, which is eight meters. This gives us, uh, this gives us 16 times April 25 So that's 1 32 meters squared per second squared. So you take the square root of that and you get V not equals 11.5 meters per second. That's partner and part B. Same technique. Uh, first resolving forces in the white direction you get again. That end is mg co signed data, no surprises there. And then in the white and the ex direction. Ah, this time you have f Secare is acting in the positive extraction, right? It's moving upwards just positive. Plus our be another mike of minors. The began the horizontal component gravity Angie, sign data. That's equal to mass times acceleration on DH. So and so this is Mrs K Tons mg cosign Taito minus and she signed data equals mass times acceleration. And so you're masses cancel and you notice that a setbacks is very similar to part A. The only difference is a negative sign. Sure. Eso this becomes a becomes ci times Mrs K Co sign data minus sign data. Uh, as opposed to both of these terms being negative part and And so this is, uh this is just 9.8 times point to be coast 37 minus sign 37 and we get that acceleration in the extraction is negative. 3.55 meters per second squared. So in this case, you also have acceleration down the hill down the ramp, but with a smaller magnitude this time and final step is again the equation of motion. And so again, same thing isn't here for you, not squared is equal to two times x times X and so that would be or negative two times. So that would be negative. Two times that of three point 55 times eight meters. And so the negatives cancel. Um, and you did what you get is and you take a square area of this and you get B not equal 7.54 meters per second.

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