A small block of mass m=1 kg glides down (without friction) a circular track of radius R=2 m, starting from rest at height R. At the bottom of the track it hits a massless relaxed spring with spring constant k= 11 N/m, which starts to be compressed as the block continues to move horizontally. Note that we assume no energy loss during this “collision". There is friction between the block and the horizontal surface, and it is not uniform. As a function of distance, the friction coefficient varies like μ(x)=αx, with α= 1.0 m−1. Assume for simplicity that static and dynamic friction coefficients are the same, and use g=10 m/s2. (See figure)
(a) What is the maximal distance x1 that the block moves horizontally away from the track at x=0? (in meters)
x1=
unanswered
(b) What time t1 does it take for the block to travel between x=0 (relaxed spring) and x=x1 (block at first stop)? (in seconds)
t1=
I assume that you found the speed of the block as it left the track and hit the spring from conservation of energy
Call it Vi = sqrt(2gh)= sqrt(2gR)
the Ke of the block when it hits the spring is then
Ke = (1/2)m Vi^2
the retarding force of friction is
Ff = m g mu
where mu = 1x
so Ff = m g 1 x
the retarding force of the spring is
Fs = k x
in compressing distance d the total work done against friction and spring is
Integral 0 to d of (mg+k)x dx
which is
(1/2)(mg+k)d^2
so in the end
(1/2)m Vi^2 = (1/2)(mg+k)d^2
Hi Damon, so we have to find 'd' from this equation? and how to find time?
@Damon: thx
@asha:
solve for d:
(1/2)m Vi^2 = (1/2)(mg+k)d^2
-> d=sqrt(m Vi^2/(mg*alpha+k))
but I also don't know the time
"d=sqrt(m Vi^2/(mg*alpha+k))"
whats that alpha in your solution?
shouldn't it be
d=sqrt(m Vi^2/(mg*+k))?
Think of it as a harmonic oscillaton, and then find k', which does not equal k.
After that, you can find the period( T) of the motion, what's the relation between t(x_0->x_1) and T.
I used T=2*pi*sqrt(m/k) and I divided it by 2, but it was wrong. Please BBB can you tell more about it?
I also tried this to find k':
(1) when spring is compressed
1/2mv^2=µmgx+1/2kx^2
(2) when spring returns back
1/2kx^2=µmgx
hence from (1) & (2) :
1/2mv^2=2µmgx
x=(1/4)* v^2/µg
and
1/2kx=µmg
k=2µmg/x
since µ=alpha*x I used µ=alpha*x1 (x1 found in part a)
I then used t1= half a period
t1 = pi*sqrt(m/k')
But it's wrong.
I lost the game :) but I want to understand what I did wrong and what is the good way to do it.
I also had c) wrong... based on my not good at all b) :)
Thank for the explanation!