A block of mass 7.82 is sliding to the right along a frictionless surface at v=

Question

A block of mass 7.82 is sliding to the right along a frictionless surface at v= 1.50 m/s. A rough patch of length of 2.74m is at point B. This patch is characterized by µs= 0.47 and µk- 0.12. The block slides back and forth between the two ramps.

A. How many times will the block corss the rough patch?

B. Where, exactly, will the block finally stop? (be specific)

C. Which ramp (left or right) does the block go up last? why?

D. On the last trip up the ramp from part C, how high up does the block go?

Here is a link to the diagram:
file:///Users/emdoucette/Desktop/Screen%20Shot%202013-11-21%20at%2010.37.51%20AM.png

Please help, be as detailed as possible!!

A block of mass 7.82 is sliding to the right along a frictionless surface at v= 1.50 m/s. A rough patch of length of 2.74m is at point B. This patch is characterized by A^ mu s= 0.47 and A^ mu k- 0.12. The block slides back and forth between the two ramps. How many times will the block cross the rough patch? Where, exactly, will the block finally stop? (be specific) Which ramp (left or right) does the block go up last? why? On the last trip up the ramp from part C, how high up does the block go? note: In real life, a block would get caught on the sharp angles shown. Assume for this problem that the block traverses the angles smoothly.

Explanation / Answer

initila kinetic energy at the bottom, ki = 0.5*m*v^2 + m*g*h

= 0.5*7.82*1.5^2 + 7.82*9.8*3.15

= 250.2

workdone by friction when the block slides on rough surface,

Wf = -mue_k*m*g*d

wf = -0.12*7.82*9.8*2.74

= -25.198 J
A)
the block crosses the rough surface 9 times

B)

mue_k*m*g*x = 250.2 - 9*25.198

0.12*7.82*9.8*x = 23.418

x = 23.418/(0.12*7.82*9.8)

x = 2.546 m (from right end of the rough surface)

C) right ramp

D) m*g*H = 250.2 - 9*25.198

7.82*9.8*H = 23.418

H = 23.418/(7.82*9.8)

H = 0.306 m ( on right ramp)

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