Posted by Anonymous on .
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=

PHYSICS(HELP) 
Anonymous,
x_1=sqrt(2gRm/(mg*alpha+k))

PHYSICS(HELP) 
Damon,
When does this exam finally end?

PHYSICS(HELP) 
Anonymous,
when some one tells us what t_1 is

PHYSICS(HELP) 
fighter,
plz tell about t_1

PHYSICS(HELP) 
anonymous,
this exam ends tommorow at midnight

PHYSICS(HELP) 
asd,
how to find alpha?

PHYSICS(HELP) 
Hawk,
alpha is 1.5 m^1

PHYSICS(HELP) 
Hawk,
how can we find t???

PHYSICS(HELP) 
fighter,
SOMEONE PLEASE TELL THE FORMULA FOR t_1

PHYSICS(HELP) 
Anonymous,
t_1 = pi / (2*w) where w = sqrt((k+alpha*m*g)/m)

PHYSICS(HELP) 
Greco,
how did you find alpha?

PHYSICS(HELP) 
Hawk,
Greco, alpha is 1.5 m^1
or I am wrong??? 
PHYSICS(HELP) 
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