Urgent Please Help. Physics Question?

Question

A block is pushed against the spring with spring constant 7.6 kN/m (located on the left-hand side of the track) and compresses the spring a distance 5.1 cm from its equilibrium position (as shown in figure below).

The block starts at rest, is accelerated by the compressed spring, and slides across a frictionless track except for a small rough area on a horizontal section of the track (as shown in the figure below).

It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.8 m/s^2

PLEASE REFER TO THIS DIAGRAM :

http://img391.imageshack.us/my.php?image=thisdiagramfu7.jpg

(A) What is the speed of v of the block when it leaves the track ? Answer in units of m/s.

(B) What is the total speed of the block when it hits the ground? Answer in units of m/s.

Answer 1

spring transfers elastic PE = increases block's KE

0.5 kx^2 = 0.5 m[v^2 -0]

v^2 = (k/m) x^2

v = (k/m)^1/2 * x >>>>>>> (1)

block will travel with (v) till patch begins, then r****d due to friction force = Fc = - u *mg

let its final speed be vx >

vx^2 = v^2 - 2 [Fc/m] s

vx^2 = (k/m) x^2 - 2 [u* g] s

vx^2 = (7600/0.414) (0.051)^2 - 2 [0.5*9.8] 1.5

vx = 5.75 m/s

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at top > gets fired horizontally vy =0 (initial)

vy-f^2 = vy^2 + 2 gh = 0 =2gh = 2*9.8*1.8 = 35.28

vy-f = final = 5.94 m/s

v-o^2 = vx^2 (remains same) + vy-f^2

v-o^2 = 33.05 + 35.28=68.33

v-o = hitting total speed = 8.267 m/s