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Hockey puck (mass 3.5 kg) leaves the players stick with speed 16 m/s and slides on the ice before coming to rest- The coefficient of friction between the puck and t...

Question

Hockey puck (mass 3.5 kg) leaves the players stick with speed 16 m/s and slides on the ice before coming to rest- The coefficient of friction between the puck and the ice is 0.6. How far will the puck slide after leaving the players stick?

hockey puck (mass 3.5 kg) leaves the players stick with speed 16 m/s and slides on the ice before coming to rest- The coefficient of friction between the puck and the ice is 0.6. How far will the puck slide after leaving the players stick?



Answers

A hockey puck leaves a player's stick with a speed of $9.9 \mathrm{~m} / \mathrm{s}$ and slides $32.0 \mathrm{~m}$ before coming to rest. Find the coefficient of friction between the puck and the ice.

Hi here. This is the horizontal, rough, icy surface over which this is the hockey pod which is moving towards right with initial speed. We I, which has given us 35 meter per second. If the mass off this hockey pop is M, then its weight MGI will be acting vertically downward. And the normal reaction exerted by this surface over the hockey puck is n where this n is equal to m g. The weight off the hockey pa No, as the hockey puck is moving towards right so there will be, ah, force of friction acting between the surface and the hockey puck, which will be resisting the motion off this hockey park due to which finally, this hockey park will come toe rest. The force of friction acting on it is given as mu times off. Its weight means mu into energy, which, as for Newton's second off motion, may be equated with mass the product off mass with acceleration. Or we can say it retardation. So as it is retardation, hence we can right it like retardation offered by the I see rough. I see surface. He's given us A is equal to minus new Gee as it is retardation Soul We are putting a negative sign Vivid Now the distance traveled before coming to rest to distance. Troubled by this hockey puck is given us 95 m hence using third equation off motion which says we f square is equal toe re I square plus two a s for we have this zero for Vitus is square off 35 meter per second plus two and this is minus mu into G and four G. This is 9.8 meter per second square in tow s which is 95 meters So this expression can building is two times off New in 29.8 into 95 meter square per second squared is equal toe 35 square meters were per secondly square So canceling this meter square for secondly square this new comes out Toby Square off 35 divided by two times off 9.8 in 2 95. So finally, this immune coefficient of kinetic friction between the icy surface and hockey puck comes out Toby 0.66 This is the answer forgiven problem. Thank you

What is it? Well, everyone, this problem, whereas to find how far a hockey puck it is going to slide even that it has a kinetic friction coefficient with the ground on which is sliding off 0.5 We're told that the puck has an initial velocity or initial speed of 5 m per second and we want Thio. Let it roll away until it's stuff, so its final velocity or final speed is going to be 0 m per seconds. Additionally, there is no change in height, so we can conveniently set the irritation potential to be a to reference value which we can set to be zero. So we can set the height to be at 0 m, which then, since potential energy is given by mass times, do irritation, acceleration times the height. If y is zero. This is always gonna be zero for for this particular problem. And the last piece of the puzzle that we need is what is the energy going to transform into. So the the energy is going to transfer into the work done by friction in stopping or absorbing essentially the kinetic energy that the podcast. So if you think about the free body diagram of the puck. This is what it looks like, so there's a there's it has a it has a mass. So it's got some weight and it's not moving in the vertical direction. There's in normal force on it, which is equivalent to the weight in magnitude at least and associated with that normal force. There's going to be a kinetic friction uh, force, right. So the friction force is equal to, you know, the, um, kinetic friction coefficients from UK times n where n is the factor is that is the normal vector. And so the normal director, since it equal it is equal to the weight is going to be mass times g. So summing it all up the magnitude off the kinetic friction forces going to beam UK times MGI and the work done by this force is the product of this force with the displacement vector. And so this placement rector is pointing in the same direction. Or, sir, is it pointing opposite the direction off the force in this case. But the point is that the magnitude can be taken into account really easily if you know where the energy goes, So let me just talk about the magnitude here. So the work done by fiction is going to be, um, f k that it would be. And so writing out with the magnitude of this thing is it's going to be mu k Times mg, which is just the kinetic friction force times the distance that the puck roles. And this is essentially, uh, it's the same thing is the thermal energy that we have calculated another problems. So what do we want to do? What we want to set the initial energy off the puck, equal to the final energy of the puck. And so, as we've said, the potential energy Zahra zero, because we've said everything to be in the reference value and there's no change in height. And we wanted to final kinetic energy to be also zero because we want to put to stop. So what we're left with is that the kinetic energy, the initial kinetic energy of the book is going to be equal to be worked on by friction or is going to be equal to dio thermal energy generated as it stops. So putting in the expressions for these for these, uh, quantities. We find that a half times the mass of the puck B I squared sequel the U K M G D. Where the masses cancel as you can see and then rearranging for D you find that the distance that the pluck slides is going to be equal to V. I squared, divided by two times the coefficient of candidate fiction times to gravitational acceleration and putting in all the values. You find that this distance is equal to 25.5 m and so this is how far this fuck is going to slide.

Question 35 states that a hockey book leaves a player stick with the speed of 9.9 meters per second and slides 32 meters before coming to rest. Find the coefficient of friction between the puck and the ice. So this is the scenario we're dealing with we're trying to solve for the because of its moving media is all for the coefficient of kinetic friction. So we know that given our scenario, he's just draw a diagram. If the puck is being hit to the right hand, signed, uh, the block. So this is our force. The block, the puck Sorry will be slowing down. So acceleration is in this direction. But since it from moving to the rate, we also have a force of connect friction f left K opposing our motion as well. So we need a sulfur acceleration then. So, based on our givens velocity, the COMAC equation best used because we're getting a distance would be the's where Nichols be not squared, plus to a D minus 2 88 minus because we are slowing down. Of course. You know, the puck initially started finally goes to rest. You know, the final velocity has to be zero and we can solve for acceleration in this equation to meet me, not squared over two D and of course, has the negative because again focus slowing down. Given initial velocity of 9.9 meters per second his whole term. It's weird and I might two times 32 meters. You can find that the acceleration other puck in this time frame is negative 1.53 meters per second squared. So maybe eventually actually little this as a force there's no force being applied to anymore. It's just the velocity is pokey to the right. So the force must be opposing emotion. And since it comes rest, if you call this motion direction X, of course we have gravity which liberalised in terms of everybody diagram This is the force here was slowing it down. And we have the normal for us. The net force has to be zero because it is slowing down when it comes to rest. You know that the and they plus f K has to be zero fiction forces in U. K times the normal force ever trying to solve from UK. That means you can move our acceleration to the option. Science O M times bigger native has a and so are coefficient. Kinetic friction is native M. A is already negative. So ever since I said that will end with a positive divide Better known more for us was just, you know, the weight of the puck. I m over g r mass terms cancel out, which is very fortunate because we don't have We're not giving mass. So we have a negative negative acceleration value Negative 1.53 units per second squared, divided by G, 9.8 meters per second squared our units cancel out our negatives become positive and we find that our coefficient of connect friction in this scenario to to see that everything is 0.16

Yeah. Here first we write the kinetic equation. That is the square mine issues critical to two years. Here we is the final velocity you is the initial velocity and A is vaccination. An ex is the displacement. Since the fire velocity of the buck is equal to zero. The above equation is written as zero plus you square equal to two heirs. Or by rearranging it. We can fight it as equal to minus musical by two years. And we consider it as a equation one. Now we write the expression for coefficient of kinetic friction from the result of exercise 67 a. It is muche equal to minus A. G by -8 by G. And we consider it as the equation too. Hear music is the coefficient of hieratic friction and reuse the acceleration due to gravity. Now we substitute 5.0 m per second form you and 20 m for six meter for us in the equation. Want to find A. So by substituting, we get equal to minus 5.0 m per second hole square, divided by two, multiplied by 20 m. And from here we get equal to minus 0.25 0.625 m/s squared. Now substitute minus 0.625 m per second squared for a and 9.8 m per second squared posey in the equation to to find me. Okay, so by substituting the value, we get muche equal to minus minus 0.625 m per second squared, divided by 9.8 m per second squared, which is equal to 0.0 64 Therefore the coefficient of kinetic friction between ice and the park is 0.064. So this is a complete solution. Please go through this. Thank you.


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