Qualification > Sciences

Springs and momentum. 1 question CIE M/j 06 5(c)

(1/2) > >>

missbeautiful789:
<a href="http://tinypic.com?ref=2qtl1lk" target="_blank"><img src="http://i52.tinypic.com/2qtl1lk.jpg" border="0" alt="Image and video hosting by TinyPic">[/url]


M/j 06 5(c)
Please could someone help me with this question, im not sure which approach to use. Much appreciated  ;D

Deadly_king:
I take it that you found the solution to part (b)

Work done by spring = 1/2 k ( x22 -  x12)

c) You should find out the gain in strain energy of spring using the equation above. But you need to convert k in Nm-1 first. Hence k = 1600Nm-1

Gain in energy strain energy for both springs = 1/2(1600)(0.062 - 0.0452) + 1/2(1600)(0.032 - 0.0452) = 0.36 J

Strain energy gets converted into kinetic energy of trolley.
K.E = 1/2mv2 = 0.36

Taking m as 0.85kg ----> V = 0.92 ms-1

Hope it helps :)

missbeautiful789:
I used the formula from b.
my problem is:
I found that the work don to the right is 1.26J
And the work done to the left is -0.9J
so shouldn't the total work done on the mass be 1.26-(-0.9) to the right
so instead of 0.36, why isn't it 2.16J

Deadly_king:

--- Quote from: missbeautiful789 on October 24, 2010, 11:43:14 am ---I used the formula from b.
my problem is:
I found that the work don to the right is 1.26J
And the work done to the left is -0.9J
so shouldn't the total work done on the mass be 1.26-(-0.9) to the right
so instead of 0.36, why isn't it 2.16J



--- End quote ---

Hehe.........my dear missbeautiful789

Energy is a scalar quantity. So it cannot be negative ;)

missbeautiful789:
Doh.. :o
thats what you get for blindly following formulas

+Rep DK

Navigation

[0] Message Index

[#] Next page