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Cinsb HancomcntEueninenelenment Heein Date: [W17/2018 12 0E.00 AM Due Date 10/28/2018 11.59.00 (119) Froblem " mnnte 2kg E sliding down ineline (Vi = 15 ns He ...

Question

Cinsb HancomcntEueninenelenment Heein Date: [W17/2018 12 0E.00 AM Due Date 10/28/2018 11.59.00 (119) Froblem " mnnte 2kg E sliding down ineline (Vi = 15 ns He = 0.2) the left. mILSZ Mz I0 kg hange Lmom mas sless . frictionless pullcy shown_ rises Ax =0.1 slides down the slopeHow fast Mz rising in TIs?Grde SunrLar} ledenma uenin

Cinsb Hancomcnt Eueninenelenment Heein Date: [W17/2018 12 0E.00 AM Due Date 10/28/2018 11.59.00 (119) Froblem " mnnte 2kg E sliding down ineline (Vi = 15 ns He = 0.2) the left. mILSZ Mz I0 kg hange Lmom mas sless . frictionless pullcy shown_ rises Ax =0.1 slides down the slope How fast Mz rising in TIs? Grde SunrLar} ledenma uenin



Answers

$\bullet$$\bullet$ A toboggan approaches a snowy hill moving at 11.0 $\mathrm{m} / \mathrm{s}$ .
The coefficients of static and kinetic friction between the
snow and the tobogan are 0.40 and 0.30 , respectively, and the
hill slopes upward at $40.0^{\circ}$ above the horizontal. Find the
acceleration of the tobogan (a) as it is going up the hill and
(b) after it has reached its highest point and is sliding down
the hill.

This question about the concept of the chin. So there are two case in one case the objectives moving up to the hill And in 2nd case the objectives moving down to the head. So for the first part, let's struggle. Free world diagram of the object when it it is moving up to the hill. So the for such acting on the object is the mass, the weight of the object. That is the question. The mass into the preparation, execution and G. And the normal reaction force that is FN. And since the object is moving up to the hill, the friction force, well let down to the hill. That is uh milk air times. FN. Now let's throw the production of the way perpendicular to the plane and along the plane, so perpendicular to the plane is MG cost center. And along the plane is MG. Scientific. And let's say the vertical to the plane is posted. Works and perpendicular to the plane is positive by No, since the block is not moving along the perpendicular to the plane so we can write the submission of all forces along the Y direction zero Are the normal reaction fourth fn minus MG cost figure As a covering 20. The normal reaction for the chinese mg cost too. They said this is a question. Okay. Now for the second case the submission of all foods along the uh plane of the hill is equivalent to the mass into the acceleration. And the forces around the hell is I'm the scientist to minus and minus milk F. In. That is Milky MG cost center and that is equivalent to them into a and from this the insulation Yes minus Gov Science theater plus. Okay pass theater. So this is the acceleration. So the magnitude of the acceleration is G scientist er Plus milk A or seek. Mhm. And the direction is down to the and now in the second case B. So when the bloggers moving down to the hill okay the friction force will change your direction and it will act upward to the hell there's no kevin. Okay. And all forces X along the Same as the previous one. So the again the all forces along the X. Direction is equivalent to uh the mass into acceleration. And again the force along the X. We are assuming particular supposed to direction. So uh vertical forces UK MG cost data Minister delivered forces MG Scientific to. And that is the covering to M. A. All the acceleration is minus Jill Scientific to minus. Okay frosty down. Uh The acceleration is my uh G. Off Scientific data minus Milky Cost data and the direction is down. So the help now for part a. Their substitute and find out the values the acceleration is g s 9.8. We just post 2nd square And signed three days. Sign off 40° Plus milk 0.3 and cause 40 degrees are the acceleration. For the first cases 8.55 made us for a second square, and for the second case the acceleration is 9.8 m/s sq And sign off 40° -0.3. Also 40 degrees, or the X solution is 4.05 meters per seconds when the blog come down to this hell?

Question was 71 from the given situation. When doubt, block and one will fall by a high touch, then it will lose its gravitational potential energy and one G at. And this loss in gravitational potential energy is equal toe the kinetic energy off the system off two blocks, which is half and one plus. And to in tow. We lost you off the system, so stop shooting the values. Um, one is to Katie in 29.8 and falling distances to meter. This is equal tohave into four K g last to K G into the square. So solving this for we we quarto three points 61 meter per second. This is the velocity off the system.

In this problem, we have a to bargain with a girl and ball boy both writing on it, starting from rest at the top of a slope at point A and reaching some velocity at point B at the bottom of the slope, at which point the boy falls off the toboggan and it moves on with only the girl. So we wish to calculate the velocity of the bargain after the boy has fallen off. So let's first use eat conservation off energy, and we'll use this to calculate he the speed of the toboggan at the bottom of the slope. So the conservation of energy, if we use the lowest point b two b, the data they win the toboggan and the writer is appoint A. They are 3 m above the data, so conservation of in details us that the kinetic energy, plus the potential energy initially at the top of the slope must equal to the kinetic energy on the gravitational potential energy at the bottom of the slope. So initially the starts from rest they have no Connecticut energy, so that zero the potential energy is the mess. The total mess is due to the two Bogan, the girl and boy, which is 10 plus 14 plus 45 kg times G, which is nine 0.81 times. They hide above the data 3 m and this is equal to the final kinetic energy of the system, which is a half times a total mess. Again can plus 40 plus 45. This is at the bottom of the slope at point B, where the velocity is VB and that's squared and the gravitational potential energy. That point is obviously zero, so we only have one unknown here, which is the velocity off the system, the boy girl and to bargain at the bottom of the slope. So this is VB and it's seven point 672 meters per second. So all we do is rearrange the above equation. Now let's apply the relative distance formula. Sorry, the relative velocity formula, rather now from relative velocities, the relative velocity of the falling boy with respect to the two Bogan, we'll call it vb slash t, and that's 2 m a second. So the velocity off the boy is equal to the velocity off the toboggan, plus the relative velocity of the boy to the toboggan. So we're taking in this case, they're at the bottom of the slope and from a diagram will take the left as positive. And so the velocity of the boy is equal to the velocity of the bargain minus to. So he moves off with the relative velocity of 2 m per second to the right. Now we'll call this equation. One will use this later. Now, next will apply the conservation off linear momentum. Uh huh. Wow, If we consider the toboggan and the writers as a system, then the impulsive force caused by the pushes internal to the system and therefore to cancel out. So as a result, the linear momentum is conserved along the X axis. So the linear momentum at the bottom of the slope is the total mass off the two riders and to bargain times the common velocity V B must equal to the situation after when the boy falls off the to bargain with his mass and B and velocity V B plus the momentum off the toboggan and the girl empty plus m g times the velocity of the to bargain VT. So again we'll take the left direction as positive and we get that the total momentum before the boy falls off his 10 last 40 plus 45. And we calculated the velocity to be seven point 672 at point B. And this was equal to the mass. Off the boy 45 times is velocity when he falls off Peter Bergen. Yes, the massive toboggan and go, which is 10 plus 40 times VT. So we have another equation here with two unknowns V V and VI ti. So we call this equation to So now we have two equations with the same too unknowns. So we can solve equations one and two simultaneously. And if we saw them simultaneously, we confined VT with which we wish to solve for And if we do that, we get the speed of to begin after the boy falls off is equal to eight 0.62 meters the second

All right, I've drawn a diagram of the problem. Um There's a block sliding toward a wedge and the wedges also sliding towards the block. The block slides up the wedge to a height of 037 m and everything is frictionless. So we can use both conservation of momentum and conservation of energy here. So mass of the block trying to right here, time speed of the block, plus the mass of the wedge, times speed of the wedge is going to equal the combined system because in the end the blocks just going to be sitting on the wedge, um speed of the block wedge system. Also, kinetic energy initially is going to be one half mess the block via the block squared plus one half mass of the wedge V. Of the wedge squared. And that equals one half mass of the combined block plus wedge V of the block, and the wedge squared plus the mass of the block times gravity times the height that it rose to. So that's a potential energy. So what I did is I went to Dez most dot com and I grafted, I put in numbers for both equations. The first equation is in black and the second equation is in red. Um What I further did was I left X be the mass of the wedge. And why is the velocity of the block wedge system? I then see that obviously the mass which is X. Has to be positive, so it's over here now. Um Why the velocity of the block wedge system doesn't necessarily have to be positive. However, they do need to intersect those two curves do need to intersect. And so this is a reasonable answer. I'm wondering whether there would be a reasonable answer all the way out here. But it looks to me like they're never going to intersect, it's like there assam toting to the same number, but there not going to intersect with each other. And I could cry, I'm sure I could uh re arrange that and see if there is a horizontal assam tote uh and see if the horizontal acetone is the same number and if it is then they would never intercept intersect down there. So the point is there is the answer. Um The mass of the block. No, no, no. The master. The wedge is 1.5 kilograms and the velocity of the block wedge system is going to be zero point for eight meters per second. And that makes sense because the block is coming at a much higher speed than the wedge. And so it may cause the wedge to reverse direction. So, thank you for watching.


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