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A crate of fruit with a mass of 32.0 kg and a specificheat capacity of 3800 J/(kgâ‹…K)J/(kgâ‹…K) slides7.00 m down a ramp inclined at an angle of36.0 degr...

Question

A crate of fruit with a mass of 32.0 kg and a specificheat capacity of 3800 J/(kgâ‹…K)J/(kgâ‹…K) slides7.00 m down a ramp inclined at an angle of36.0 degrees below the horizontal.If the crate was at rest at the top of the incline and has aspeed of 2.55 m/sm/s at the bottom, how muchwork WfWf was done on the crate by friction?Use 9.81 m/s2m for the acceleration due to gravity andexpress your answer in joules.

A crate of fruit with a mass of 32.0 kg and a specific heat capacity of 3800 J/(kgâ‹…K)J/(kgâ‹…K) slides 7.00 m down a ramp inclined at an angle of 36.0 degrees below the horizontal. If the crate was at rest at the top of the incline and has a speed of 2.55 m/sm/s at the bottom, how much work WfWf was done on the crate by friction? Use 9.81 m/s2m for the acceleration due to gravity and express your answer in joules.



Answers

A crate of fruit with mass 35.0 $\mathrm{kg}$ and specific heat 3650 $\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}$ slides down a ramp inclined at $36.9^{\circ}$ below the horizontal. The ramp is 8.00 $\mathrm{m}$ long. (a) If the crate was at rest at the top of the incline and has a speed of 2.50 $\mathrm{m} / \mathrm{s}$ at the bottom, how much work was done on the crate by friction? (b) If an amount of heat equal to the magnitude of the work done by friction goes into the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change?

In this question. We have a create of fruit of mass, 35 kg and a specific heat of 3650 sliding down a ramp inclined at 36.9 degrees. The ramp is 8 m long and were asked to calculate how much work is done on the Great by friction if it slides down from rest and has a final speed of 2.5 m per second. And then we are asked to calculate the temperature change. If the work done by friction, uh, is equal to an amount of heat that is transferred into the Great. So the idea behind this question is that the crate slides down the incline. Um, the speed increases, but there is a force of friction. S o The speed is not increasing as much as it would have if it was just sliding down this ramp without friction. Friction is taking out some energy from the crate as it slides down the ramp. So since friction is taking energy out, um, one of the forms of energy that that could be converted into is heat. So this question is asking us to suppose that all of that work done by friction. All that energy taken out of the crate, Uh, in terms of its kinetic energy or it's mechanical energy. Ah is converted completely into, um, heat energy. So that's the basic premise of this question. Let's go ahead and jump into the calculations. So in the first part were asked to calculate the work done on the crate by friction. Now there's a couple different ways that you can calculate work so we can calculate work using force and displacement. Or we can calculate work using three overall change in energy. So, um, I'm going to use the second way today. Both are options here, but I just think that calculating the change in energy here is just a bit easier. So the work done by non conservative forces or a K A friction is going to be equal to three overall change and energy, and that's going to be composed of the change in kinetic energy, plus the change in potential energy. So the change input in kinetic energy is going to be one half MV F squared, minus one half MV I squared. But of course, this thing is starting from rest so this term is going to be zero and then the change in potential energy can be written as M G. Delta H. Now, we're not given Delta H specifically in the question, but the ramp is 8 m long and the angle is 36.9 degrees. So we can easily find the heights by calculating the height of this triangle. Um, so it will be Delta H. So the Delta H here is going to be sign 36 0.9 times eight. Just the hype oddness. Okay, so now we can go ahead and plug everything in. We've got a mass of 35 kg. The final speed is 2.5 m per second is going to be squared plus 35 times g times the height, which if we solve this, we get a height change in height off four points. 8 m. Now, we do want to consider the change in height as being negative, because as this great slides down the ramp, we're experiencing a negative change in potential energy, meaning the potential energy is decreasing. So we need to consider the height as negative. So we get that change in potential energy to be negative as well. So once we go ahead and plug all of those numbers in, we're going to get in overall amount of work. Negative 1.54 times 10 to the three jewels. So the change in the total mechanical energy here is 1.54 times 10 to the three jewels. The negative indicates that we're losing that energy. And so that is the work done by friction. That's the work, um, or the energy taken out by friction. Okay, for the second parts, um, were asked to calculate the heat if all of that work done by friction. Um, sorry. We're asked to calculate the temperature change if all of that work done by friction goes into heat. So we've got a formula that relates heat and temperature change, and so we can easily rearrange this to find Delta T. So that's going to be cute over EMC. And for Q, we're going to use that amount of work that we just found. So that's gonna be 1.54 times 10 to the three jewels, and then the rest is just found in the question. So 35 kg and then we can go ahead and fill in that specific heat. And once you put that into a calculator, you get about 1.2 times 10 to the minus two degrees Calvin or degrees Celsius. So the overall change in temperature that would be experienced by the great here if all of the work done by friction goes into heat is actually very, very small. So this is the final answer, Thio Part B and then the final answer to party.

The coefficient of friction doesn't really matter in the sense that here, um, the force vector is parallel to the displacement vector. So they're giving us essentially the angle of the ramp, Bates. And they're giving us the vertical distance of the ramp. But here we are applying the 202 125 noon force parallel to the ramp. So we're going to take were simply finding if we're taking the X axis to be along the ramp, we can simply find the work done in the X axis. This would be the work done equaling the force, the magnitude of the force times the distance. In this case, this would be the full length of the ramp, not just the vertical distance. So this would be Equalling to the force multiplied by the height of the ramp, divided by sine of data being the angle of the incline. And so we find we can substitute the work equaling the force. 225 Newtons multiplied by 1.15 meters. This would be divided by sign of 30.0 degrees. And we find that then the work being done is equaling two, approximately 518 jewels. That is the end of the solution. Thank you for watching

In this particular case, what is happening is we're trying to find work done by gravitational force. So remember that gravitational force is in this direction, which is mg and the distance travelled is in this direction equals and and here is angle Tater. So what I've done by the gravitational force equals give to MGI the force times the distance l times the co sign angle between Ellen mg with Vico sign off 1 20 degree. So that is eight times 9.8 times five times. Course I anyone 20 degree equals negative 1 96 Jews walked on by the tension. The tension is in this direction with some angle Alfa. So work done by the tension equals attention Times l times course I enough. The angle between them which is given by 18 degree. You know, they tend to this 120 Newton, then is five meter times who signed 18 degree so that comes out to be 570 Jewell Now the normal forces over here and the angle between name is 90 degree. So walk done by the normal force equals zero for Part B. We have friction along the surface, so for these, we had to find what is the normal force. So the normal force thus g sign 18 degree Equus, MGI, and this equals MGI. Go sign Tedder. Well, today study. Really? So from here we can find. And it was m Geico sine theta minus G sign 18 degree Putting all the values in like this to be 30 degree, 30 degree on DA Mass to be We're doing part B now, so that's masses 50 gram or 0.5 kilogram. Using these, we find the normal force to be 30 warned eight new done. So in the second guess work done against the friction force is given by the coefficient off magnetic fiction times 30.8 New done. That is mule in times the work done, which is fight meter and that comes out to be 30.8. Julen, no, Uh, this number waas wrong. Ah, Okay. The kinetic friction waas 0.25 So this is 0.25 So this number comes out to be 38 0.5 June

And this problem. We have an 85 kilogram and pushing a crate four meters along 20 degrees slope for him. He exerts a force of 500 Newtons parallel to the ramp and him and the Creator are moving at a constant speed against Rick. We want to calculate what the total work done is by this man. And for doing this we have to consider two things. One is the work that he does in moving Crate. The second is the work that he does in moving his body against gravity. So the work that he does and moving crate will be equal to our applied force on this crate old supplied by our displacement and then the co sign of the angle between the direction of our applied force and a displacement. We're told that our forces parallel to the ramp, so that means it's in the same direction as it is. This displacement in our angle is zero So f g co sign of zero, just one. So the work that he does and moving the crate is just that Applied Force 500 news multiplied by our four meters. Now in moving his body against gravity. So against gravity, we have to recall that the work done on a system is a transfer of energy. And with there being no change in his kinetic energy, he's starting and stopping at the same speed we're told is moving at a constant speed. His only change in energy is gonna be a change in potential energy in this gravitational setting. So we know that the work that he does in moving his body up along this ramp will be his change in energy, which is gonna be his change in potential energy. And we know that our potential energy that change is gonna be our mg Delta H. So what is dealt age? Well, we know that all right, Ran is tilted up 20 degrees from the horizontal in the ramp itself. He's moving it up four meters. So this doubt that age or he's starting here, stopping here that's gonna be equal to four. Sign 20. So we have this work against gravity. It's gonna be MG times for sign 20. This means that his network in moving from the bottom of the ramp to the top of the ramp, I was gonna be the work that he does on the crate, plus the work that he does against gravity moving his body. For this 1st 1 we had forced times, the displacement. And now we're told that he is. And 85 kilogram man said that Emmett is 85. Gravity 9.8. Yeah, for sign one. And if we plug this in, we get so three significant figures 3140 jewels.


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