ok, should i square root those two.
I have also tried:
0.5kx^2=0.5mv^2+mgh+Fd
0.5kx^2=304.08+4272-4264
0.5kx^2=312.08
*2
kx^2=624.16
so
kx=1066.97
kx^2=624.16
equal equations for k, rearrange fro x:
x=624.16/1066.97
x=0.58, so k=1839.6, or is this method also wrong
What I did to get that is use the sin rule to get the vertical heigh (vh) which was:
vh=sin(21)*8/sin90=2.87
which i then put in the GPE formula to get:
GPE=mgh
GPE=152.04*9.8*2.87
GPE=4276.28
Which I then inserted into:
kx^2=mv^2+mgh-Fd
kx^2=608.16+4276.28-4264= 620.44
so
kx^2= 620.44...
I think I have calculated it right, by calculating the vertical height, which gave me an answer of 1839.7, can you please tell me if this is the correct answer, or if you got something different
From my calculations I have got:
kx=1066.97
kx^2=14336.72
Which if I then both equal to k, I get
1066.97/x=14336.72/x^2
Which then I multiply with x, to eliminate one side, and then rearrange to get;
x=14336.72/1066.97
x=13.44
Which is the wrong answer as x cannot be higher than 8, if 8...
Ok, I think I know what you are trying to tell me, and I think I may have done the method you are trying to tell me, but I always end up stuck in the last part, with the rearranging:
force on spring= force on slope+friction
F=kx=mgsin(theta)+friction
F=1066.97
Elastic spring energy= kinetic...
Ok, thank you for helping me first of all. Basically what I am trying to do is figure out the spring constant by obtaining the force, and the distance the spring moved and the force exerted on the spring
First I have tried finding the force on the spring by calculating force on the slope, as...
I am not looking for an answer, just a method that gives me a result close to 1871.88. I have tried this for hours and when I entered that value it says it is wrong but by a small amount. Please, can someone help me as soon as possible, the assignment is due at 9pm tonight.
1. Homework...