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Topic: **What is case method teaching****Question:**
This is the question:
Use energy methods to calculate the speed of the 6.00-kg block after it has descended 1.50 m.?
Consider a system where the rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the 8.00-kg block and the tabletop is uk = 0.250. The blocks are released from rest.
This is the answer:
energy lost due to friction = friction force* displacement
= 0.25*8*g*1.5
= 29.43 J , take g as 9.81
change in potential energy of the system = mgh
= 6*g*1.5 +8*g*0
=88.29 J
( 8 kg block move in horizontal plane so there is no change in its potential energy)
assume the velocity of the 6.0 kg block is v, 8.0 kg block also has the same velocity if it is a simple pulley system
change in the kinetic energy = 0.5*m*v^2
= 0.5*6*v^2 + 0.5*8*v^2
= 7*v^2
according to principle of energy conservation,
7*v^2 + 29.43 =88.29
v= 2.9 m/s
What I don't understand is, why is work due to friction not negative in this question? Because friction is working against the direction its moving , thanks
this may be a stupid question but what is the constant

June 16, 2019 / By Austin

You have ignored signs entirely and intuited your way through the problem, and got the right answer. Intuition is good, but it either works or it doesn't. In a complicated problem, you might not "just get it". If you know the rules you can navigate your way through it step by step. "energy lost due to friction = friction force* displacement = 0.25*8*g*1.5 = 29.43 J , take g as 9.81" Since you said energy lost, you should recognize that as negative work. If you go step by step, it comes out that way too. Friction force is in the opposite direction of displacement. W = Fd cos theta theta = 180 degrees cos theta = -1 W = (0.25*8*g)(1.5)(-1) = -29.43 J "change in potential energy of the system = mgh = 6*g*1.5 +8*g*0 =88.29 J" Change in potential energy here is negative. That should always be the case for something falling. When in doubt, determine potential energy after minus potential energy before. In this case, we may say the before energy is for a height 1.5m, and the after energy is for a height 0m. delta PE = PE after - PE before = 0 - 88.29 = -88.29J "change in the kinetic energy = 0.5*m*v^2 = 0.5*6*v^2 + 0.5*8*v^2 = 7*v^2" This is correct, but to be precise, you should do kinetic energy after minus kinetic energy before. It just happens that kinetic energy before was 0 because the system was at rest, and so that doesn't affect the result. delta KE = KE after - KE before = 7v^2 - 0 = 7v^2 "7*v^2 + 29.43 =88.29" Why did you choose to arrange the equation in this way? You have |delta KE| + |W| = |delta PE| | | means absolute value to indicate that you've used positive signs for each quantity. Did somebody teach you that formula, or did you make it up because it just seemed right? It gets you the right answer. The standard formula is delta KE + delta PE = W Since in this case delta PE and W are negative, and delta KE is positive, this equals |delta KE| - |delta PE| = -|W| which can be rearranged to your formula. But if the signs were different, the formula in terms of positive terms would not be the same, and you couldn't use the formula you used. It is more reliable to use the standard formula, and use the correct sign convention for each term. You may then use your intuitive method to check that the result makes sense.

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Did you like the answer? We found more questions related to the topic: **What is case method teaching**

You have ignored signs entirely and intuited your way through the problem, and got the right answer. Intuition is good, but it either works or it doesn't. In a complicated problem, you might not "just get it". If you know the rules you can navigate your way through it step by step. "energy lost due to friction = friction force* displacement = 0.25*8*g*1.5 = 29.43 J , take g as 9.81" Since you said energy lost, you should recognize that as negative work. If you go step by step, it comes out that way too. Friction force is in the opposite direction of displacement. W = Fd cos theta theta = 180 degrees cos theta = -1 W = (0.25*8*g)(1.5)(-1) = -29.43 J "change in potential energy of the system = mgh = 6*g*1.5 +8*g*0 =88.29 J" Change in potential energy here is negative. That should always be the case for something falling. When in doubt, determine potential energy after minus potential energy before. In this case, we may say the before energy is for a height 1.5m, and the after energy is for a height 0m. delta PE = PE after - PE before = 0 - 88.29 = -88.29J "change in the kinetic energy = 0.5*m*v^2 = 0.5*6*v^2 + 0.5*8*v^2 = 7*v^2" This is correct, but to be precise, you should do kinetic energy after minus kinetic energy before. It just happens that kinetic energy before was 0 because the system was at rest, and so that doesn't affect the result. delta KE = KE after - KE before = 7v^2 - 0 = 7v^2 "7*v^2 + 29.43 =88.29" Why did you choose to arrange the equation in this way? You have |delta KE| + |W| = |delta PE| | | means absolute value to indicate that you've used positive signs for each quantity. Did somebody teach you that formula, or did you make it up because it just seemed right? It gets you the right answer. The standard formula is delta KE + delta PE = W Since in this case delta PE and W are negative, and delta KE is positive, this equals |delta KE| - |delta PE| = -|W| which can be rearranged to your formula. But if the signs were different, the formula in terms of positive terms would not be the same, and you couldn't use the formula you used. It is more reliable to use the standard formula, and use the correct sign convention for each term. You may then use your intuitive method to check that the result makes sense.

work is positive if it increases the energy of the system.. I think you may have done it incorrectly just looking at the change in potential energy that seems incorrect ... if the 6 kg block drops 1.5 m its potential is LOWER at the bottom change in potential = potential final - potential initial potential initial will be higher so this MUST BE NEGATIVE.. ALso think of force as a vector when the force is in the direction of motion the work is positive.. so with a spring pushing an object the direction of motion is in the same direction as the force so the work done by the spring is positive.. if there was friction involved .. then the work done by the friction would be NEGATIVE because it slowing the object or removing kinetic energy .. I think you may have done something incorrect..

work is positive if it increases the energy of the system.. I think you may have done it incorrectly just looking at the change in potential energy that seems incorrect ... if the 6 kg block drops 1.5 m its potential is LOWER at the bottom change in potential = potential final - potential initial potential initial will be higher so this MUST BE NEGATIVE.. ALso think of force as a vector when the force is in the direction of motion the work is positive.. so with a spring pushing an object the direction of motion is in the same direction as the force so the work done by the spring is positive.. if there was friction involved .. then the work done by the friction would be NEGATIVE because it slowing the object or removing kinetic energy .. I think you may have done something incorrect..

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Work/Energy are scalars and direction of motion is not important. Some teach Work as negative. I don't like that approach and argue against it. Mathematically (Conservation of Energy): PE + KE + Work = constant If all 3 terms are +, then the standard algebra processes yield correct results. If Work were - and you moved it to the right side of the =, then it would be added to the constant (Remember crossing the = Work would change sign and become + which would be a violation of Conservation of Energy)

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Negative thought pattern is not the correct term. What you have is usually called racing thoughts. Try sitting with your back straight, your hands together at your waist, looking downward. Count your breathing one to ten, if you loose count or go past ten go back to one, start the count over. Do this for about ten minutes everyday, it'll make a difference.

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