A bicylist of mass 75 kg (including the bicycle) can coast dow...

Question

A bicylist of mass 75 kg (including the bicycle) can coast down a 5.0° hill at a steady speed of 12 km/h. Pumping hard, the cyclist can descend the hill at a speed of 32 km/h. Using the same power, at what speed can the cyclist climb the same hill? Assume the force of friction is proportional to the square of the speed v; that is, F𝒻ᵣ = bv² , where b is a constant.

Accepted Answer

Welcome back. Everyone in this problem, a skateboarder with a total mass of 60 kg can skateboard Don a ramp inclined at four degrees to the horizontal at a steady speed of eight kilometers per hour by pushing off more forcefully. The skateboarder can increase their speed to 16 kilometers per hour. Using the same effort at what speed can the skateboarder climb up the ramp? Assume the forces, sorry, assume the force of air resistance and rolling friction is proportional to the square of the speed V. That is the frictional force equals KV squared where K is a constant for our enters A says it's 10 kilometers per hour. B 13 kilometers per hour, C 17 kilometers per hour and D 21 kilometers per hour. Now here, ok. In this problem, we basically have three scenarios. So let's first try to draw it and understand all three. Let's start with the first one. Ok. Now it says our skateboarder is skating at a steady speed of eight kilometers per hour. What does that really look like? Well, if I were to try to draw that here, so imagine that we have our incline. Ok. Think I can bring the soap a bit more. And on our incline, we have our skater because our skater right, and our skater is going at a steady speed of eight kilometers per hour. So let's call that speed V one. Ok. And at this speed, if we think about the forces acting on our skater, we know he's inclined at four degrees. Let's call that angle theta. And for our skater, we know his weight is going to be acting vertically downwards. So that's MG. While relative to the incline, he'll have a normal force acting acting perpendicularly to the surface. So I think I can move this up a bit more to make space for that. And so he'll have his normal force acting in this direction. OK. Let's call that N and then he'll have the Y component of his weight acting perpendicular to the incline. OK. So let's call the angle between the vertical and the Y component of his weight theta. So we know that his weight, the Y component of his weight will be MG cost theta. OK? Because it's adjacent to the angle and the X component of his weight will also be acting downward as MG sine theta. OK. No, again, as we said, we're assuming our Y and X axis are relative to the incline. OK? And for the skater, then he'll have a frictional force that's going against his direction. OK. So let's call that fr one. So in the first scenario, these are all the forces that are acting. No, in our second scenario, it says that our skateboarder pushes off and increases his speed or her speed to 16 kilometers per hour. So essentially their speed is two times V one. So what forces are acting at this point? Well, no, at V two being equal to 16 kilometers per hour. Let me just put in brackets up here. V two equals two V one. OK? It's essentially the same diagram. They, they, they're still skating down the ramp so I can just copy and paste this here. And but now in this diagram, the only different forces that are acting is that because he pushes off, he now has a force that we can label as FP one, he pushes off with a power P one and the frictional force acting is going to be a different frictional force that we can refer to as FR two. OK? So let's call this F I two. So these are the forces acting. In the second scenario, we still have his weight, having an X and Y component and acting vertically downward. Now, what about our third scenario here? Well, in our third scenario, we want to find out if our skateboarder uses the same effort at what speed they can climb up the ramp. OK. Now, in that case, then we don't know what that speed is. So let me just highlight that in yellow here. That's the speed we're trying to find. So let's call that speed V three. OK? So now at V three, which we don't know again, the diagram is going to look the same, but now we're gonna have some uh some different directions. So if I were to bring that over here, OK. Well, then no, get rid of this here. No, we have another force EFP two which in this case, like we said, is using the same effort. So it's still FP one basically. So we still have that effort. FP Well, no, it's going in a, in a, in a different direction. Ok. So now if we think about it, then we have FP two which again, as we said is FP one, which is in a different direction going upwards, ok? And we have our frictional force acting against our skateboarder, ok. So our frictional force is now going down this way, let's call that FR three because now as we said, he's going up the ramp. So it's basically similar forces, but in the opposite direction and we are trying to figure out what that speed is going up that ramp, ok. Now, if we're going to figure that out, ok. Uh So far, we already know that our frictional force is proportional to the square of speed, that is our ruling friction, ok? And we have a formula for that, we know V two equals to V one. We also know that our mass, the mass of our skateboarder is 60 kg. And we're trying to find V three, V three is what we don't know. OK. Where, as we said earlier, we defined V one and V two. Now, what can help us? Well, we can apply Newton's second law of motion to figure out what's happening in all three of our free body diagrams. And then we can see where we go from there. OK? Because recall that by Newton's second law of motion, it tells us that the sum of the forces is directly proportional to the acceleration by the formula that says the sum of the forces equals the mass multiplied by the acceleration. We can assume that the cyclist is moving at a steady speed for each case. So that means the sum of the forces is going to be equal to zero because there's going to be no net acceleration. So we can resolve those forces along the X axis if we consider the direction of motion along the X axis as positive. So let's go ahead and try to do that. So we're going to find the sum of four force is in the X direction for cases 12 and three, let's start with case one. OK. So for case one, we know that some of the forces in the X direction is going to be equal to zero. OK. And if we look at our diagram, then that means negative fr one, a first frictional force going opposite to MG sine theta is going to be equal to zero. That means fr one equals MG sine theta. OK. And remember earlier, we said that our rolling friction is KV squared. So four V one fr one is really KV one squared. So KV one squared equals MG sine theta. So let's keep that in mind for no, let's go to our second case. Now at V two, what are our forces acting here? Well, again, we know that they're moving at a steady speed, sum of forces equals zero. And if we look at all of the forces, then we know negative F two acting in this direction is acting opposite to MG sine theta and FP one. So plus MG sine theta plus FP one is going to be equal to zero. That means then that FP one equals Fr two plus MG sine theta or sorry, fr two minus mg sine theta. OK? Cause in other words, we move both of them to the other side. And no, we can simplify if we try to express our frictional force in terms of MG sign the, how can we do that? Well, earlier we learned that KV one squared equals MG sine theta. And we know that V two is two times V one. So if we take our second frictional force, OK, we know that our second frictional force is going to be equal to KV two squared, which by definition is going to be K multiplied by two V one or squared. So fr two actually equals two V one squared, which is four V one squared multiplying by K or four KV one squared. Remember KV one squared equals mg sine theta. So if we replace our second frictional force with uh our definition here of KV one squared. And what we're really seeing is that it equals four mg sine theta. So what does that mean for us? Well, no, we can substitute that value into our formula for our four FP one. So FP one is gonna be equal to four mg sine theta. OK, minus mg sine theta, which means it equals three mg sine theta. OK. So that's our force. FP one for that effort expressed in terms of MG and theta. We are the power associated with FP one which is P one equals its force multiplied by the speed. So that's FP one multiplied by V two which since we know FP one is three mg sine theta, that's really three MG sine theta multiplied by V two. So now that we have our first FP one expressed in terms of MG sign the and its power. Let's move on to our third diagram to see if we can get any further with the information that we have here. Now, first again, let's resolve our forces like we did in the other two. So now we're finding the sum of the forces in the X direction. We know it equals zero. And based on our diagram or skateboard or skateboarder is now moving up the ramp and since he's moving up the ramp, then FP two. OK. Positive minus MG sine theta minus our third frictional force. They're both moving downwards, all of that equals zero. OK. So here we have this, we just like in our previous problem, the power P two associated with FP two is going to be equal to FP two multiplied by V three. But what do we know about that power? Well, if we look back in our problem, it says that our skateboarder uses the same effort to go down with that speed of 16 kilometers per hour to climb up the ramp. So that tells us, then if we make a note here, we know from that, that P two must be equal to P one. OK. So since P two equals P one, that means P two, FP two, V three is going to be equal to P one which we found to be three MG sign theaters multiplied by V two. Now, why are we doing this? Well, now we'll be able to express our force FP two in terms of MG theta and V two and V three where V three is the only unknown. So that means then that FP two equals three MG sine theta multiplied by V two divided by V three. Now, what else can we glean here? Because no, we have an expression for FP two, we know FP two in terms of known variables except uh the variable that we're trying to find how about or third frictional force. Well, again, we could apply um what we know about the rolling friction and, or the rolling friction being proportional to the square of speed. But that's not really going to help us much because we don't know what V three is. However, we know that we're using the same effort and with the same effort, the frictional force to the square of the speed is going to be proportional. In other words, with the same effort. OK. Then, or third frictional force two, the square of the third speed is going to be equal to the first frictional force to the square of the first speed. Now again, how is that going to help us? Well, that's pretty useful. Let me put the line here. That's pretty useful because we know that our first frictional force equals MG sine theta. So we can solve for our third frictional force using that idea because no, if we replace uh fr our first frictional force with MG sine theta and multiply both sides by V three squared. OK, then we should find that our third frictional force equals MG sine theta. OK. Our first frictional force multiplied by V three squared, divided by V one squared or V three divided by V one R squared. So now we have our third frictional force in terms of MG theta V one and V three. So since we have these, these are some very important results. So since we have FP two and FR three, in terms of non variables, we can substitute it into our expression for our third case. And when we do that, then we, it, it will probably make it easier to isolate these variables and solve. So let's do that. So we can solve them for V three. So solving for V three, solving for V three by substituting FB two and fr three. OK. Then let's take FP 23, MG sine theta multiplied by V two divided by V three K minus mg sign theater minus fr three which we now know is MG sine theta multiplied by V three divided by V one R squared. OK. So all of this that we have no equal to zero. Notice that all three terms have MG sine theta in common. So we can factor MG sine theta out to get mg sine theta. Well, you outside of factoring. So we can divide the entire equation by mg sine theater. I think that would be easier. So if we divide all the terms by MG sine theta, OK. OK. Divide the entire equation by MG sign theater. Basically, then we're left with three multiplied by V two, divided by V three. OK. Minus mg sine theta divided by MG sine theta. That's one minus mg sine theta divided by MG sine theta leaves us with one again. So that's V three squared divided by V two squared. So all of that equals zero again, we are solving for V three here, we have it in the denominator. So we can just try to get rid of the denominator throughout our entire uh expression. So we can multiply the entire equation by V by V one squared. OK. Or this should have been the one squared here. Let me fix that quickly. This is V one square kick. My apologies for that. So we can multiply through by V one squared multiplied by V three. OK? Because no, if we multiply our entire equation by that, looking at our first term V three cancels V three. And that leaves us with three, V one squared, V 21 multiplied by V one squared. V three is V one squared V three. And for our third term V three squared multiplied by V three equals V three. Q. Well, V one squared would cancel out in the denominator. So all of that is equal to zero. Now, we know the values are V one and V two. So let's start to substitute because no, that means it's going to be equal to three multiplied by V one squared. So that's eight kilometers per hour squared, multiplied by V two, that's 16 kilometers per hour. OK? Minus V one squared. Again, that's eight kilometers per hour or squared multiplied by V three minus V three cubed. So all of that equals zero. No, we want to solve for V three. So we'll probably need to make negative V three Q positive. So we can multiply again our entire equation by a negative one. OK. So when we multiply the whole thing by negative one, well, here three multiplied by eight squared, multiplied by 16 equals 3072. OK? So that's gonna be negative 3072 uh eight kilometers per hour squared, multiplied by V 38 squared is 64. So that's 64 V three. And I have that minus V three equals zero. So it's sorry, it's gonna be plus V three equals zero because this is gonna become a positive here. OK? Likewise is this one and it's always good to make sure your signs are correct as you go along. You know what I should have done probably first. I should have made V three, cued the subject first, but it's OK. We've sorted out our sign. OK? And no, since that's the case, we can rearrange this to say V three cubed plus 64 V three minus 3072 equals zero. No, at this point to figure out V three, we just need to solve our equation by a cubic method because we're solving for the solving from V three Q I know you can do that. However you want, you know, you could use a math solver, you could put it on a graph. You could use algebraic methods. I'm going to leave that up to you, but however, you do it, OK? So however you solve so solving by an appropriate method, OK, then you should find that you get V three to finally be equal to 13.075 kilometers per hour. Notice that all of our answers were into two significant figures. So if we do the same, we get V three to be equal to 13 kilometers per hour. Thus, the skateboarder has to climb up the same ramp at a speed of 13 kilometers per hour at the same effort. If we look back on our answer choices, that means B is the correct answer. Thanks for watching everyone. I hope this video helped.

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