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Let point A be the where she is at rest (the start)

let point B be when she is at the end of the slope, let her speed be v ms^-1

let point C be when she is about to go up the second slope, and let her speed here be u ms^-1

let point D be when she comes to rest again.

let x be the distance of the first slope

take g to be 9.8 ms^-2

from points AB:

Gain in KE: [itex] 0.5\times v^2 \times 40 = 20v^2 [/itex]

Loss in GPE: [itex] xsin25 \times 9.8 \times 40 [/itex]

therefore by the work energy principle:

[itex] 392xsin(25) - 20v^2 = 18x [/itex]

[itex] x(392sin(25) - 18) = 20v^2 [/itex]

from points BC:

Loss in KE: [itex] 0.5\times u^2 \times 40 - 0.5 \times v^2 \times 40 = 20u^2 - 20v^2 [/itex]

as there is no change in GPE, by the work energy principle: [itex] 20u^2 - 20v^2 = 15\times 18 [/itex]

[itex] 20u^2 = 270 + 20v^2 [/itex]

from points CD:

Loss in KE: [itex] 20u^2 - 0 = 20u^2 [/itex]

gain in GPE: [itex] 25sin(6) \times 9.8 \times 40 = 9800sin(6) [/itex]

by the work energy principle:

[itex] 9800sin(6) - 20u^2 = 25\times 18 [/itex]

[itex] 9800sin(6) - 450 - (270 + 20v^2) = 0 [/itex]

[itex] \frac{9800sin(6) - 720}{392sin(25) - 18} = x = 1.92... [/itex]

however this is wrong and the answer is 11.8m

help :(?