A Spring pushing an object up a 25 cm slope

[Zero]

Homework Statement

A box is initially at rest in the following system. The box is placed next to the spring and the spring (along with the box) is compressed 30.0 cm. The spring constant of the spring is 150 N/m, and the mass of the box is 540 grams.
a. Calculate the speed of the box at position B (assume energy has not been lost due to friction).

Relevant Equations

Since I know energy is lost due to friction would I use the formula ff= U (Fn) where Fn= 5.292 since that would be the gravity action on the object and normal for and gravity have to be equal. If I use the same logic would Ff= -Ek or no???

f= kx
= 150 n/m (.3m)
= 45n

Ug= 1/2 kx^2
= 1/2 (150) (.3)^2
= 75 (.09)
= 6.75 j

Ek= 1/2mv^2
6.75= 1/2 (.54) (v^2)
6.75= .27 (v^2)
25= v^2
5= v1

 

[jonbovi]

Dear Zero,

I understand, you can neglect the energy lost due to friction. Thus, to solve the problem you can use the law law of conservation of energy.

You need to know the speed of the box when the spring pushes it. Then, the different of potential and kinetic energy will give you the speed in the top (point B).

It means the final kinetic energy will be $T = 1/2mv_B^2 = mgh-1/2mv_A²$. Now you need the speed at A. That's easy, the potential energy of the spring will be the kinetic energy of the box. In other words

$$1/2kx^2 = 1/2mv_A^2$$

Now, you only have to substitute $v_A$ in the previous equation to solve the speed at B

 

[Zero]

Dear Zero,

I understand, you can neglect the energy lost due to friction. Thus, to solve the problem you can use the law law of conservation of energy.

You need to know the speed of the box when the spring pushes it. Then, the different of potential and kinetic energy will give you the speed in the top (point B).

It means the final kinetic energy will be $T = 1/2mv_B^2 = mgh-1/2mv_A²$. Now you need the speed at A. That's easy, the potential energy of the spring will be the kinetic energy of the box. In other words

$$1/2kx^2 = 1/2mv_A^2$$

Now, you only have to substitute $v_A$ in the previous equation to solve the speed at B

I don't know if I'm doing something wrong but I keep getting a negative number when I am left with v^2 which I can't square root. 1/2mv^2= .27 v^2 and mgh-1/2mv^2= -5.427 so therefore i get v^2=-20.1 which i can't square root. Can i just switch which one comes first then the answer would be positive and i can square root from there?

 

[jonbovi]

That's true, it is negative. That means that is not possible to reach the B point

 

[Zero]

That's true, it is negative. That means that is not possible to reach the B point

So what your saying is that its a trap, if so then I wasted hours on nothing. Thanx man really appreciate the help.

 

[jonbovi]

Oh, No sorry, I realize I messed up!

The right difference of energy is $1/2mv_B^2=1/2mv_A^2-mgh$

And this changes the sign. Sorry, my fault!

 

[Zero]

Oh, No sorry, I realize I messed up!

The right difference of energy is $1/2mv_B^2=1/2mv_A^2-mgh$

And this changes the sign. Sorry, my fault!

its all good man i should have also picked up on this too. Thank you I got the answer.