Compression of a spring with friction

Some parts of the problem text are washed out, and the solution is out of focus. I don't think you want any guesswork on our part, so you should give us something more readable if you need help.

 

PE+KE=W_f+PE_s

mgh + 1/2 mv²= mgμd + 1/2 kx²
Check for zero PE reference level.
Gravitational force is conservative, only levels are relevant not the path between the levels.

 

Yes.

 

Can you show your working.
Check the unit of spring constant k.

 

I get a very similar answer(≈7.7m).

 

mgh + 0.5mv²=mgμd+0.5ky²
2x9.8x10 +0.5x2x7x7=2x9.8x0.6x20 + 0.5x0.2/0.01 x y²
196+49 =235.2 +10xy²
9.8=10xy²

y²= 0.98
y=1m

 

Why is h = 10 m and v = 7m/s? I computed the velocity at C to be 15.7m/s and the motion is at a height of 30m. (I took the reference of zero potential at ground level).

 

As i mentioned on previous reply, it is a conservative force. It is not he path but the different of height that determines the ΔPE. So the ΔPE is height between A and C.
Conservation of energy. The kinetic energy remain the same. As in projectile, the horizontal velocity remains constant.

From point A to C no energy dissipated.

add: If level A equal to level C, do you need to take zero reference at ground level?

 

Ah of course! But I still don't see why the velocity at C is 7m/s. Can you show this mathematically? I have tried but get values greater than this.

 

There are too many steps here that cause errors. Eliminate all the intermediate steps. Start at A and end at E. [itex]
(KE + PE)_{initial- at- A} + W_f = (KE + PE)_{final- at -E}[/itex]
Work done by friction is numerically a negative value.

 

Oh heck you have a good understanding of work energy concepts, so it's just math now for the 2nd part and using g = 10 , then

1/2(2)(7)^2 + (2)(10)(40) + 0 - .6(2)(10)(20) = 0 + 2(10)(30) + 1/2(20)(x^2)

From which x = 1 m.

Which looks like what azizlwl got using a slightly different approach between a different set of points. You should confirm this on your own, as there are lots of ways to skin a cat, as they say.

Oh sorry I didn't pay attention to part A I guess that's solved already.

 

Can anyone enlighten me as to why the velocity at C is 7m/s? Many thanks

 

You should confirm it is correct on your own.....I didn't take the time pouring through your pages of calcs to see why you arrived at a different answer...maybe too many steps and a number got transposed somewhere... or maybe our answer is wrong you know.....