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Find an expression for the maximum height yMAX the block goes.?
A block of mass m is pushed against a spring with negligible mass, and force constant k, compressing it an amount d1. After the block is released, it moves along a horizontal surface and then up a frictionless incline (the angle doesn't matter). The horizontal surface is also frictionless, except for a small rough patch of length d2 where the kinetic friction coefficient is uk (mu-sub-k). Find an expression for the maximum height yMAX (y-sub-MAX) the block goes.
I used the equation (1/2)kx^2 = mgLsin(theta) and substituted d1 for x and yMAX for L to get yMAX = [ (1/2)k(d1)^2 ]/[ mgsin(theta) ] but I don't think that's right. Help? I'm not sure where uk and d2 are supposed to play a role.
1 Answer
- civil_av8rLv 77 years agoFavorite Answer
Change in energy of the system = -friction work = -friction*d2
Change in Spring PE + Change in gravity PE = -friction*d2
We can ignore KE because it starts from 0 and ends a 0 at the top of the incline
1/2*k*(xf^2 - xi^2) + m*g*(yf - yi) = -friction*d2
The angle doesn't matter for the gravitational PE
Assume xf = 0, yi = 0, and from the problem statement, xi = d1
-1/2*k*d1^2 + m*g*yf = -friction*d2
From the definition of frictoin
friction = uk*N
Since the surface is flat, N = m*g
friction = uk*m*g
-1/2*k*d1^2 + m*g*yf = -uk*m*g*d2
Solve for yf
yf = -uk*d2 + 1/2*k*d1^2 / (m*g)
