1. A spring with a force constant k hangs vertically. A block of mass m is attached to the un-stretched spring and allowed to fall from rest. Find an expression for the maximum distance the block falls before it begins moving upward. 2. Now..from what i knw F=-kx (k=spring constant, x= extension) and i think it should be done like this: F=-kx mg=-kx x=-mg/k but i guess i did it wrong because my teacher did it in another very complicated way which i cannot understand. PLease see if you can help me. Solution: let da maximum distance be y. Total Mechanical evergy= total kinetic + total potential energy 0 = (1/2)mv^2 - mgy + (1/2)ky^2 and from here y was evaluated giving a result of y= 2mg/k My question is ..what is the total potential energy and y? What is the value of V, is it 0 , if yes, then y? and why is the total mechanical energy 0?
Kinetic and potential energy hello everyone.. please help me solve this problem if you can Q. Two blocks are attached to a light string that passes over a mass-less, frictionless pulley. The two blocks have masses m1= 3 kg and m2 = 5 kg and are initially at rest. Find the speed of either block when the heavier one falls a distance h = 4 m. solution: total kinetic energy: 0.5m1v^2 + 0.5m2v^2 Total potential energy: m1gh-m2gh Total energy=0 = K.E + P.E from this equation, speed,v is calculated confusion: why is the total energy 0? m2 loses P.E and that is why it is -ve m1 gains P.E, so it is +ve ..i understand that...bt what about the kinetic energy? they both gain K.E is that y they are positive? please help me.
hey phys guru..thanx..its worked..x= 2mg/k ...bt is it like a formula that For maximum extension>>>> mv^2=kx^2 isnt there a way to derive that formula?
hey...cant we derive it like this : work done = 0.5mv^2 work done by a stretched spring :0.5kx^2 so..if they are equal... mv^2 = kx^2 right ?
what i dont understand about the 1st problem is what my teacher did with the potential and kinetic energy...I completely dont get it...can u help me with that plz ?
You solved a different problem: How far does a block stretch a spring when gently hung from it. But in the assigned problem, the block is dropped from its initial position. To solve this, use conservation of energy. The total mechanical energy is conserved. Total mechanical energy is the sum of KE, gravitational PE, and spring PE. Y is the distance that the block moves down. Your teacher chose to measure gravitational PE using the initial position as a reference. At the initial position, all three terms of the mechanical energy equal zero (it's not moving, the spring is unstretched, and the gravitational PE is zero), so the total is zero. As the block lowers, gravitational PE lowers while spring PE rises. There are two values for y which make the total mechanical energy = 0. The initial position (y = 0) is one; solve for the other.
I still dont get why mechanical energy should be zero? i get that initially it was zero...bt after the block moves down, why shall it be zero ?
intial total mechanical energy should equal to final total mechanical energy...is that y the fine is 0?
how do i knw when evergy is conserved. A force is conservative if the total work it does on a particle is 0 when the particle moves around any closed path returning to its initial position. examples are gravity and electromagnetic. But how am i supposed to know when this work done is 0? like in the case of the hanging masses m1 n m2 stated earlier, the particles do not return to their initial position, how can i say that energy is conserved. I think i understood the term wrong. What does it actually mean-"energy is conserved"? Initial energy equal final enery,and hence the change is 0 ? please help me out..im in a great dilemna
Since there are no dissipative (non-conservative) forces, mechanical energy is conserved. Right. That's just a general statement about conservative forces. If you move an object around, then return it to its origin, the work done by gravity is zero: gravity is conservative. Same thing for the spring; spring forces are conservative. Exactly.
Thanks a lot Doc Al! it really helped me...i have a few more confusion and problems..please stay around to help me out.
It seems DocAl has some personal rivalry with me or what... For me the expression is completely relevant to the problem.In fact i have solved all my high end physics problems with the same expression and got full marks. I apologize if i am too rude or offensive but the fact remains that the expression i just provided has been in Indian text books since ages for the same category of problems.