Dilemma About Energy : Can You EXPLAIN? [closed]  PROBLEM 1 : A 2.14 kg block is dropped from a height of 43.6 cm onto a spring of force constant k = 18.6 N/cm as shown in fig:12-19.find the maximum distance the spring will be compressed.

solution :::: Let the spring get compressed a distance x. If the object fell from a height h= 0.436 m . then conservation of energy gives

0.5 * k * (x^2) = mg(x+h) { note that we calculate "mgx" here } now if we put all the values we will get x = 0.111 meter

PROBLEM 2 : figure 12-16 shows a 7.94 kg stone resting on a spring . the spring is compressed 10.2 cm by the stone . the stone is pushed down an additional 28.6 cm and released. how much potential energy is stored in the spring just before the stone is released? and how high above this new (lowest) position will the stone rise?

solution ::: if x distance spring compressed because of mg ( m is the mass of block) we can get force constant k = F/x = mg/x = (7.94)(9.81)/0.012 = 764 N/m

after the stone is pushed down an additional 28.6 cm x = (0.286 + 0.102) meter

SO potential energy stored in the spring just before the stone is released

0.5 * k * (x^2) = 0.5 * 764 * (0.286 + 0.102)^2 = 57.5 J

if h is height from the lowest position to the highest it will rise,using conservation of energy we can say mgh = 57.5 J so h = 0.738 meter

note that we didnt use "mgx" here !!

now my question is why didnt we use mgx in second problem?? shouldnt we solve it like the first one??

problems collected from "Physics" by halliday,resnick,krane volume 1 5th edition this book isnt used as textbook in our country so this isnt a homework problem. im reading the book for my own pleasure.

closed as off-topic by Danu, Norbert Schuch, ACuriousMind♦, John Rennie, Sebastian RieseDec 11 '15 at 15:51

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In the first:

The energy of the mass till it reaches the height of uncompressed length of the spring: $E_i=mg(h+l)$

where $h$ is the height of the mass dropped from the uncompressed length of the spring

Final energy after compression: $E_f=\frac{1}{2}kx^2+mg(l-x)$

Equating $E_i$ and $E_f$

$\frac{1}{2}kx^2=mg(h+x)$

In second prob

$E_i=\frac{1}{2}k(x_1+x_2)^2+mg(l-x_1-x_2)$

$E_f=mg(h'+l)$

So, $mg(h'+x_1+x_2)=\frac{1}{2}k(x_1+x_2)^2$

$h'$ is the height of the mass will reach from the height of uncompressed length of the spring

This quantity $"h'+x_1+x_2"$ is basically taken as $"h"$ in your solution

• sorry, i don't get it. what do you mean by 'l' in mg(h+l)? – tahsin Dec 11 '15 at 18:12
• uncompressed length of the spring – Oswald Dec 11 '15 at 18:13

In the first picture, and before you push down the spring loading mass in the second picture, you have two forces with two corresponding energies, the force of the spring $F_s=k(x-x_0)$ and $F_g=-mg$, energies $E_s={1\over 2}k(x-x_0)^2$, and gravity $E_g=-mgx$, and they are equal and in opposite directions at equillibrium.

$F_s+F_g=0$

So their magnitudes are equal.

The same can't be said about when you push down the extra distance on the spring in the second problem. You are applying a force, lets call it $F_A$, say it goes in the same direction as gravity, and now recognize that is added to the equality.

$F_s+F_g+F_A=0$

Now gravity alone doesnt describe how tightly the spring is wound to be in equillibrium, but since you can figure out the spring constant and know its displacement from the initial equilibrium length you can determine the energy stored by the spring.

• In the first picture, and before you push down the spring loading mass in the second picture, you have two forces, the force of the spring Fs=0.5*k*(x^2) and gravity Fg=−mgx, and they are equal and in opposite directions at equillibrium. Fs+Fg=0 ###### that's not correct. please put their values and you will see they aren't equal – tahsin Dec 11 '15 at 16:09
• oops, I made a typo – Skyler Dec 11 '15 at 21:33
• Those are the corresponding energies, and then I meant to put the forces next – Skyler Dec 11 '15 at 21:33
• well, even their corresponding energies are not same. please put their values and notice that. 0.5 * k * (x^2) is not equal to mgx!!! – tahsin Dec 12 '15 at 13:41
• In the second example, obviously the two energies are not the same, there is a hand or whatever else that is also pushing down the block. In the first example when you reach equillibrium there is no movement, so there is no kinetic energy term, meaning the sum of ALL your potential energies (a spring initial gravity initial and final of both), will be equal. Initially I was just outlining the directional and functional dependency. It's up to you to pick the right coordinates and make sure all your variables are consistent. – Skyler Dec 12 '15 at 22:07