Finding the spring constant and the distance of the block?
the question is too long, i couldn't write it or upload it in Yahoo uploader,
this is the link of it : http://d.top4top.net/p_86ub1h1.png
m = 4.00 kg
theta = 30
v = 1 m/s
compression of the spring by the block , x = 0.3 m
d = 5 m
rule : Fk = μ k * N
i must find Fk first to find μ k using conservation energy,
First find N,
N = m x g x cos theta
N = 4 * 9.81 * cos30 = 33.98N
then we find the force to find the spring constant
F = ma
F = 4 * 9.81 = 39.24N
KE = 0.5 mv^2
KE = 0.5 * 39.24 * 1^2 = 19.62J
PE = 0.5 kx^2 => 0.5 Fx
PE = KE
0.5 Fx = 0.5 mv^2
x = v^2 / a = 1^2 / 9.81 = 0.101m
K = f/x = 392.4
Here i am confused, our professor told us to use mechanical energy and conservation energy to find Fk ,
and the rules is
delta E = delta E mechanics + delta E thermo + delta E internal
K2 + U2 = K1 + U1
The question is long and very tricky, i solve what i can, and i not sure about it actually, if you can help me i will be very thankful.
- NCSLv 74 years agoFavorite Answer
I question a couple of your steps:
(i) F = 4 * 9.81 = 39.24N
Why do this? This is the weight of the object.
(ii) KE = 0.5 * 39.24 * 1^2 = 19.62J
This is wrong. You have put the weight in instead of the mass.
(iii) delta E = delta E mechanics + delta E thermo + delta E internal
That looks OK, but
K2 + U2 = K1 + U1
has lost something. Instead, you should wind up with
deltaE = deltaThermo = friction work = initial GPE - final SPE
since the deltaKE is zero.
So, start with the horizontal setup to find k:
spring PE becomes block KE
½kx² = ½mv² → ½ cancels
k * (0.20m)² = 4.00kg * (1.00m/s)²
k = 100 N/m
Now consider the incline problem:
initial GPE = friction work + final SPE
m*g*(5.00m + 0.30m)*sinΘ = µ*m*g*(5.00m + x)*cosΘ + ½k(0.3m)²
Plug in m, g, k and Θ and solve for µ.
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