A 2.0-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 250 N/m) which has its other end fixed. The block passes through the equilibrium position with a speed of 2.6 m/s and first comes to rest at a displacement of 0.20 m from equilibrium. What is the coefficient of kinetic friction between the block and the horizontal surface?
2007-12-06 16:38:59 · 4 answers · asked by roncho 4 in Science & Mathematics ➔ Physics
YES anilbaks? (sp) that is actually more helpful...it is so clear now...and i can use it as a template....thank you so much...
2007-12-06 17:59:48 · update #1
fixed spring.....x=0 | .........(+x) ........ O block
at time (t), x be displacement from equlibrium
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note >> restoring force is opposite to direction of movement of block, and friction also acts opposite to movement of block>> so friction adds to restoring force
-------------------- suppose block moves in +x direction at t=t
restoring force F = - k (x) = - k x (i) >>> towards - x
friction force = fc = uk * mg (- i) assisting restoring force
net force towards +x
= ma (+i) = kx (-i) + uk mg (-i)
ma = - (k x + uk mg)
dv/dt = - w^2 x - uk g >>> w^2 = k/m, uk = kinetic friction
dv = - (w^2 x - uk g) dt = - (w^2 x - uk g) dx/v
v dv = - (w^2 x - uk g) dx
integrating
0.5 v^2 = - [0.5w^2 x^2 + uk g x] + C
at x = 0, v = 2.6 m/s
0.5(2.6)^2 = C
0.5 v^2 = - [0.5w^2 x^2 + uk g x] + 0.5(2.6)^2
multiply by 2
v^2 = (2.6)^2 - [w^2 x^2 + 2uk g x] ------ (1)
at x = 0.20 m, v =0 comes to rest
0 = (2.6)^2 - [w^2* 0.20^2 + 2uk g*0.20]
(2.6)^2 = [0.04 w^2 + 0.4 uk g]
0.4 uk *g = 6.76 - 0.04 w^2 = 6.76 - 0.04*(250/2)
0.4 uk *g = 6.76 - 5 = 1.76
uk = 1.76/0.4*9.8 = 0.4489
uk = kinetic friction = 0.45
2007-12-06 17:17:39 · answer #1 · answered by anil bakshi 7 · 2⤊ 0⤋
Energy at the equilbrium position is all kinetic
1/2 m v²
Energy at the extended position is all potential
1/2 k A²
Compute them both and take the difference; the loss is the work done by friction. Set that work equal to Force x distance to get the average force of friction. That force is μmg where μ is the coefficient of kinetic friction.
2007-12-06 16:45:11 · answer #2 · answered by jgoulden 7 · 1⤊ 1⤋
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2016-10-10 11:00:55 · answer #3 · answered by ? 4 · 0⤊ 0⤋
that is a hard one
2007-12-06 16:41:14 · answer #4 · answered by Kolaid 2 · 0⤊ 1⤋