Question

A block with mass m = 2 kg lies on a frictionless surface. The block is located on a compressed
spring , as shown below. The spring is compressed 20 cm from the equilibrium position and has
spring constant of 500 N / m. After the spring is released slid mass away and after a short
slide it begins to slide up a hill with 45◦
gradient. How far up the hill will
block go before it turns ?

Answer 1

I advise you to use Conservation of Energy.

Answer 2

Yes I know, and potential energy will convert to kinetic energy as soon the spring is released and from that i have initial velocity and from that I use wp1 - wp2 = wk1 - wk2 in the hill?

Answer 3

But I havent worked with springs so much so I need help with the beginning, how to make it into a velocity

Answer 4

Potential spring energy = 0.5*k*(s)^2
Where 'k' is the spring stiffness (500 N/m)
And 's' is the distance the spring is AWAY FROM ITS EQUILIBRIUM LENGTH.
That note in caps is important, because some questions can trip you up with that.

Answer 5

Let's set the datum for the gravitational potential energy. Set it at ground level (which is usual place) to make the problem easier.
Then, we can observe the initial setting, and the intermediate setting.

Answer 6

In the initial state, there is potential spring energy. No gravitational potentiail energy, because we're at ground level. And the block has no kinetic energy. So, the only form of energy is the spring energy.

Once the spring acts on the block, I assume the spring goes back to equilibrium. It doesn't say though. So, there is no more spring potential energy. Still no gravitational potential energy. And now, the block has kinetic energy.

So, your first equation will be:
U_spring_initial = K_block
0.5*k*(s)^2 = 0.5*m*v^2

Answer 7

In the final section, you'll have the kinetic energy of the block turn into all gravitational potential energy at the highest point of the ramp. And, I'm sure you can do that. Plus, you'll have to use some trig to figure out what that actual distance of the ramp is.

Answer 8

But from the first equation: 0.5*k*(s)^2 = 0.5*m*v^2, I have to solve for v? Just to make sure!

Answer 9

Potential spring energy = 0.5*k*(s)^2. s must be (-0.2)m? so it is released 0.2 m in the positive x direction?

Answer 10

You don't have to solve for v. The energy is more important. You can find the kinetic energy since it'll be equal to the spring energy.

And for the energy of the spring, strangely enough, when using Conservation of Energy, the spring energy will always be positive. So, it doesn't matter if the spring was compressed or stretched, s will be 0.2 m. And if you square it, you'll always get a positive value, anyway.

Answer 11

But dont I need a initial velocity to find out how far up the hill it will come?

Answer 12

Nope, because you already have the kinetic energy, which equals the spring energy.
And in the second half of the problem, the kinetic energy will equal to gravitational potential energy. Don't need the velocity at all.

Answer 13

Btw, with gravitational potential energy, you'll get the distance in the y-direction the block travels, since gravity only works in the vertical direction. Then you'll have to use some trigonometry to figure out the length of the ramp the block went up.

Answer 14

so the U_spring= 10 J? and so is W_k?

Answer 15

@iPwnBunnies

Answer 16

Yes

Answer 17

U_spring=0.5*(500N/m)(0.2)^2?

Answer 18

Yes, that is correct. I forgot to square the distance.

Answer 19

oh, okey :)

Answer 20

But I think you need to help me with the second part, Cause I cant understand how I can find the hight?

Answer 21

I need the height to calculate the potential energy

Answer 22

gravitational potential energy*

Answer 23

Yes, you're looking for the height the block travels.
But as I said, the kinetic energy all turns into gravitational potential energy.
So, 10 = mgy

Answer 24

ah I see

Answer 25

And again, that 'y' will only be the vertical distance. Then, you'll need to use trig to find the distance up the ramp. I think you're good from here. c:

Answer 26

\[10=mgy => 10=2kg*9.8*y => 10=19.6N*y => 10/19.6=y?\]