The path that a block slides along is frictionless until the block reaches the section L shown in the figure below. The length L = 0.75 m begins at height h = 2.0 m on a ramp of angle θ = 30º. In that section, the coefficient of kinetic friction is 0.40. The block passes through point A with a speed of 8.0 m / s.
If the block can reach point B (where the friction ends), what is its speed there, and if it cannot, what is its greatest height above A?
[tex]K_A = U_g + K_f -f_kd [/tex]
The Attempt at a Solution
L (friction spot)= 0.75m
[tex]h_f= 2.00m [/tex]
[tex]\theta= 30^o [/tex]
[tex]\mu = 0.40 [/tex]
[tex] V_A= 8.0m/s [/tex]
I was thinking conservation of momentum...
do I take A. to be the initial V? I was thinking it would be but not sure. (continuing as if I'm correct)
[tex]K_A + U_g= U_g + K_f -f_kd [/tex]
[tex]0.5 mv_A^2 + 0 = mgd sin 30^o + 0 - \mu mgL [/tex]
Is this correct?
or does the distance for the friction have to include the angle?
and thus be
[tex]0.5 mv_A^2 + 0 = mgy sin 30^o + 0 - \mu mgLsin 30^o [/tex]
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