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A bicyclist of mass 75 $\mathrm{kg}$ (including the bicycle) can coastdown a $4.0^{\circ}$ hill at a steady speed of 12 $\mathrm{km} / \mathrm{h}$ . Pumping hard, the cyclist can descend the hill at a speed of 32 $\mathrm{km} / \mathrm{h}$ . Usingthe 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, $F_{\text { tr }}=b v^{2},$ where $b$ is a constant.

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$29 \mathrm{km} / \mathrm{h}$

Physics 101 Mechanics

Chapter 8

Conservation of Energy

Work

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

Cornell University

Rutgers, The State University of New Jersey

University of Washington

University of Winnipeg

Lectures

04:05

In physics, a conservative force is a force that is path-independent, meaning that the total work done along any path in the field is the same. In other words, the work is independent of the path taken. The only force considered in classical physics to be conservative is gravitation.

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

04:04

(III) A bicyclist can coas…

07:28

A bicyclist is coasting s…

03:57

03:00

A bicyclist is coasting st…

04:32

A bicyclist can coast down…

02:43

06:03

ssm A bicyclist is coastin…

01:55

(III) A bicyclist coasts d…

03:09

(1II) A bicyclist coasts d…

01:22

(II) How fast must a cycli…

01:57

If a bicyclist of mass 65 …

02:06

11:50

Let's get started with some free body diagrams here. So this is the downhill case. Um, and we're defining our axes such that the normal force direction upwards eyes is positive y axis as positive white and positive X is uphill. Okay, so, um ah, when the cyclist is coasting down the hill, the force of friction, Of course, gravity and normal for a second downwards and upwards has shown here. But when cyclist is coasting, our force affection is opposing, uh, his motion. And so the and then And so that's the only horizontal force that is acting opposite the direction of travel. When the cyclist is pumping, uh, pumping the brakes are been applying of pumping force. It's called a S a P one. You again have a fiction force acting opposite direction of motion. We gonna call this F How far? One. But there's a P one, which is a pumping force from downhill motion, which is acting in the negative xx. Okay, so, um, let's get started With Newton's second law resolving forces in the ex direction when you're coasting. Friction force is just equal to horizontal component off gravity. MG signed data because there's no um, men acceleration. He is, um, coasting, Um, when that's when is pumping, uh, again you resolve forcing the ex direction. But now you have f so far in one is equal to F p one plus And she signed data, right? And so we wanted. And so f p one is a quantity of interest. The pumping force, which is equal to ah far one minus mg, signed data. Now, how do we know what so far one is? We know that f eyes proportional to V squared velocity squared. Therefore, taking a ratio of f ah fr 12 f f r we take a ratio. Um, take the ratios off the two velocity. So coasting velocity is 12 kilometers per hour pumping bossy this 32 kilometers per hour. Um and so just take the ratio of those 32 over 12 and you square that because culture in proportionality, constant cancels out. So, um, f f R one. Therefore, it is 64 over nine times f f r. And what is half a far? It's Angie Scientific, as we saw here. So this is 64 over nine. Uh, whoops. Let's write it down better This is 64 over nine times en jeu signed data. Um, so therefore we have That s a P one is mg signed data. That's the common term Time 64 over nine minus one. Right. Uh, I just replaced after far wonder on DSO This is 55 over nine times mg signed data. All right. Therefore, power p one, um, is equal to FT. One times. Um, V one velocity one. Uh and, um bossy one is I lost. The one is, of course, 32 here. And so this is Ah, 55 over nine times V one mg signed data. That's P one. Okay. And we're gonna call that equation one. All right, um, when you're going uphill, um, the only difference here, the difference is that force of fiction is acting in the negative X axis now on. So when you're pumping, you're going uphill. Let's use a different color right now. Actually, you're going uphill and you're pumping. Uh, the force that you're resolving ah is you are. When you're resolving forces in the ex direction you have f p two is equal. R f you have you minus f p group's excuse me. You have f p two minus f fr to force a friction. In that case, minus mg sign. Fatal. Therefore, F p too, is f fr two plus em Jew sign, vato. Um right. And so we need to know what ffr too is and again use a ratio of ah ah, fraction of fresh and forces after far too. Over. Ffr is just, um v to which, which is something we don't know over the force from here. 12 uh, quantity squared. Therefore, ffr too is just, um mg Cynthia times V two over 12 square. All right. Ah, And so now we use the fact that power will be the same, uh, in both. For both cases, a P one will be called up you two and soapy to Izzy called, too. Uh, f p to V two. Um rate, which is equal to P one and P waters we recall is 55 over nine times V one times mg Sign data. One be wanted is 32 by the way on, so f you too is equal to 55 over nine times. We want overview too. Um, actually, you know what? Let's just write down. What we want is of 32 overview two times mg Signed data one. Ah, From here we have that F p two is fr 2% G signed data. Um, and so this is actually equal to M g side data plus f f ar, too. Which is, uh, mg signed data times. Ah, V two over 12 quantities quit. All right, um, s So therefore we have mg scientist and all the pain, all the terms. So take that comment and he signed data times 55 over nine times 32 over V two, AA minus one minus V two over 12. Ah, quantity squared is equal to zero. So what we're left to it is the falling equation V two cubed plus 12. Cubed view to minus 55 over nine times 12. Squared V one is equal to zero. Numerically, this is vey too cute. Plus one 44. The eclipse. I meant to say of these 12 squared over here. V two cubed plus 1 44 of you two squared. Uh, 1 44 V two. No squared there. Um, minus to eight. 610 is equal to zero. So you can put this into your calculator. This is 1/3 order polynomial equation. Ah, and the answer comes out to be view two equals 28.85 kilometers per hour. If you want to convert it to meters for second, go ahead. But a song as we're using the same units throughout you should be good and that is it.

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