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Question 24The energy possessed by stretched or compressed spring is the elastic potential energy:Notyet ansk eredSelect one: 0 TrueWarked out otFalseFlag quesiionO...

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Question 24The energy possessed by stretched or compressed spring is the elastic potential energy:Notyet ansk eredSelect one: 0 TrueWarked out otFalseFlag quesiionOuetonThe force of friction on body when moves over ground depends onNolytlJnsenadMattng Dut Cnormai reaction on thc bodyFla? Juastonb. b) Mass of the bodyweight of tha bodyd; d) noncabot0

Question 24 The energy possessed by stretched or compressed spring is the elastic potential energy: Notyet ansk ered Select one: 0 True Warked out ot False Flag quesiion Oueton The force of friction on body when moves over ground depends on Nolytl Jnsenad Mattng Dut C normai reaction on thc body Fla? Juaston b. b) Mass of the body weight of tha body d; d) nonc abot0



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When $2 \mathrm{~kg}$ mass hangs to a spring of length $50 \mathrm{~cm}$, the spring stretches by $2 \mathrm{~cm}$. The mass is pulled down until the length of the spring becomes $60 \mathrm{~cm}$. What is the amount of elastic energy stored in the spring in this condition, if $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ (A) $10 \mathrm{~J}$ (B) $2 \mathrm{~J}$ (C) $2.5 \mathrm{~J}$ (D) $5 \mathrm{~J}$

Hi in the given problem, the force required to stretch the spring is air. And the stretched which has taken place in the spring is X. So using the relation between the deforming for supplies and the extension taking place in the spring, it says a physical two K into X. There K is the spring factor or force constant. We can see it. So using this expression we get an expression for the spring factor or the force constant that is given by force per unit extension. And this K remains constant for a given spring. So the energy store the elastic potential energy stored in the spring will be given by U. Is equal to have okay, excess square. No, Now we want to stretch the spring two double off. Its original extension means to stretch It by two x. The force required will be equal to K. Multiplied by two X. Or we can say times of cakes and we know K. X. Was equal to F. So the force required should be trouble. Now the expression for the potential energy is falling the spring. Now you dish will be equal to half K. There will be no change in the cave, half K into X. Dash is square. And we look that X dash was twice off X. To the whole square. So it will be four times of half K. X square. And this half K X square was the initial potential energy means here it becomes four times of you. So you can say the force will be twice off the original force and potential energy story will be four times of the initial potential energy store. Hence here, our option C. Is correct. Thank you.

There we have given spring is Compressed by one cm by a force or world. Newton compression is one centimetre. Find the potential energy of the spring when it is compressed vitals and immediately so we have to find the potential energy. First of all, we know that this force will be equal to K into the attacks. So this is their taxes one centimeter. That is one and 2 20 power minus two. So the value of K will be cool too. Or indeed New England. What are we done? This is the screen constantly. Another potential energy you will be equal to have K. That's the square. Right now we're able to find this potential energy event spring. It's compressed tense and divided. So this will be called Uh huh. And took a into the centimeter Yes and limited. So we will put .1 zero square. So this is a golden 200 tingled And they into 10 to the power minus soul. That is a good to june. Now we'll check the options. So option is correct. Thank you

So here for party, we simply air going to relate hooks law negative K X to the weight of the object. This would equal mg again, the weight of the object and so X. The displacement would be equaling negative mass so negative 0.500 kilograms multiplied by the acceleration due to gravity 9.80 meters per second squared divided by the spring constant of 40.0 newtons per meter. And we find that this position is negative point one 2 to 5 meters. We can say that these spring is stretched approximately 0.123 meters and so we can say for part B, then the change in potential energy would be equally 0.500 Essentially, the mass multiplied by 9.80 meters per second squared and this is going to be multiplied by negative 0.12 to 5 meters. This is essentially the change in gravitational potential energy and so we find that the change in gravitational potential energy would be equaling negative point six 00 jewels. We can say that here the potential energy of the object decreases. That is the end of the solution. Thank you for watching

Hello students in discussion we have a particular spring that has 1/4 law F equals two minus of demon club. Excuse where these are constant. Okay so for the part A we have to calculate the potential energy you when we have given you equal to zero at X equals to zero. So we know that potential energy is given by minus of F dx. And this interrogation so minus of F D X. Which is from here will be equal to dx Q. And this D. X. And interrogation from zero to X. Okay so this is the potential energy you so you it is equal school this value deep and X cube and dx interrogation from zero to X. So we will get from here you it is equal to one by four times of the axis square. So this is the answer for the problem. Okay No for the part B we have to calculate the work required to stretch the spring from X equals to 02 X. So we have done is given by you final minus you initial so you finally is that is you are the X. That is one by four times of the axis square minus you are the X equals to geo. So this will be won by four times of demarchelier B. Zero to the power for So what then from here is one by four. The X is clear. Okay so this is the answer for the part B of the problem. Okay thank you


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