Finding Distance D Using Conservation of Energy

[patrickmoloney]

Homework Statement

A block of mass 5kg is released from rest and slides down a distance D down a smooth plane inclined at 30◦ to the horizontal. It then strikes a spring compressing it 10cm before it begins to move up the plane. If the spring constant is 980 N/m and g=9.8m/s², find D. Solve Using the Principle of Conservation of Energy.

Homework Equations

mgh=1/2kx²

The Attempt at a Solution

I drew a diaghram of the problem and found the resultant forces on the block.

Kinetic energy of the block= potential at the bottom of the spring.

mgh=1/2kx²

m=5 Kg, g=9.8 m/s² , h= Dsin30 (pythagoras theorem) , k= 980 N/m , x= 0.1m

subbing into the formula

(5)(9.8)(Dsin30)=1/2(980)(0.1)
D=0.2 m

is this the correct solution? it's an assignment so I'd like the method to be correct. Thanks in advance.

 

[voko]

You are on the right track, but. You have neglected that the block gains additional energy as it compresses the spring, because it goes lower.

 

[patrickmoloney]

how do i show this in mathematics and is D right or wrong due to this additional energy?

 

[voko]

Your basic equation $mgh = \dfrac 1 2 k x^2$ is correct. $h$ should be the height from the top position of the block to the very lowest position, where the spring is fully compressed.

 

[patrickmoloney]

the height from the top position of the block to the very lowest position is Dsin30, no? Since d=h/sin(theta)

 

[voko]

No, because the very lowest position is where the spring is fully compressed.

 

[patrickmoloney]

is h= (D+0.1)sin30?

 

[patrickmoloney]

then if h is indeed =(D+0.1)sin30 , D= 0.15 m

 

[voko]

Correct.

 

[patrickmoloney]

Apparently the answer was 0.2?? Where did I go wrong?

 

[voko]

You found previously D = 0.2 m. But now you know that instead of D, you should have used (D + 0.1 m), so D = 0.1 m.

I do not know why you think the answer should be 0.2 m.

 

[patrickmoloney]

i did use h=(D+0.1)sin30 as you said it was correct and got 0.15 m to be the answer. I did not get 0.1 m.

I thought 0.2 m due to a colleague in the class got this answer and he is better than me at applied maths and he did it a different way to me.

 

[astro2525]

49(D + 0.1)sin30 = (1/2)980*.1^2
49Dsin30 + 2.45 = 4.9
49Dsin30 = 2.45
49D = 4.9
D = .1 m

As Voko said, you find 0.2 when you don't account for the compression of the spring which contributes to the total distance. Your colleague is wrong =P.

 

[patrickmoloney]

mgh=1/2kx²

(5)(9.8)(D+0.1)sin30 = (1/2)(980)(0.1)²
Dsin30 + 0.1sin30 = (1/2)(980)(0.1)²/(5)(9.8)
Dsin30 = (1/2)(980)(0.1)²/(5)(9.8) - 0.1sin30
..oh I see. I didn't divide - 0.1sin30 by sin30 :(

schoolboy error. Guys thanks for the help.