Asked by Anonymous

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=


(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=
Answered by Anonymous
x_1=sqrt(2gRm/(mg*alpha+k))
Answered by Damon
When does this exam finally end?
Answered by Anonymous
when some one tells us what t_1 is
Answered by fighter
plz tell about t_1
Answered by anonymous
this exam ends tommorow at midnight
Answered by asd
how to find alpha?
Answered by Hawk
alpha is 1.5 m^-1
Answered by Hawk
how can we find t???
Answered by fighter
SOMEONE PLEASE TELL THE FORMULA FOR t_1
Answered by Anonymous
t_1 = pi / (2*w) where w = sqrt((k+alpha*m*g)/m)
Answered by Greco
how did you find alpha?
Answered by Hawk
Greco, alpha is 1.5 m^-1
or I am wrong???
Answered by Greco
"ì(x)=áx, with á= 1.0 m−1" so alpha is 1.0 m^-1

i got a green tick so in this problem alpha is 1.0 m^-1. In mine was
m g k R a 1 10 8 2 0.8

and the answer was
x1=1.58
t1=0.59
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