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of é We wie da) ve B Kie S orse distance tory sem boas par m . . 37° 0,8. Sunt Problem 2. A massless pulley is fixed to the t
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Answer #1

let the acceleration of the system be a m/s^2

writing the force equation for both the blocks we have

5g - T = 5a --------(1)

T - 5gsin37 - u*5*g*cos 37 = 5a ----------(2)

u = 0.25

adding both equation we have

3g = 10a

a = 3*9.8/10 = 2.94 m/s^2

T = 5(g - a) = 5*(9.8 - 2.94) = 34.3 N

c) v^2 = u^2 + 2aS

v^2 = 0 + 2*2.94*1 = 5.88

KE = 0.5*m*v^2 = 0.5*5*5.88 = 14.7 J

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