Question

A Sky block is released from rest on an inolined plane (0=209. It then slides a distance, d= 9m dann the slope before hitting a spring (K=7000 N/m). If there is friction (lek = 0.15) all along the ramp's surface, what is that maximum compression of the Spring due to ne. Block? d> we

Answer 1

given: mass m = 5kg d = 9m O= 20° K= 7000 N/m Me= 0.15 as Wwwmumi In order to find compression in spring, we have to find total energy at point B. This total energy (EB) will be transferred to the spring" Now consider the FBD of block R = Narmal reaction of incline ciis fe= friction force actiong on block (ii) mg = weight of the block. mg Casso mgsin 20 y YL

Since no mation in y. direction = z fyzo ie: R= mg Coszo' Now the frictional force fic MK.R = 0.158 mg cos 20" -0.15x5x9.8% Cos20° fic= 6.91N - Now h= AC = dSinzo = ax sin20 = 3.08 m . Now Applying Energy conservation between A & B Energy at point A-Energy loss in friction - Energy at B EA- W friction & EB (EA= mgh (patential mgh - faxd = EB Energy only frictio = work done to 5x9.8*5.08 - 6.91X9 = EB overcome friction [EB = 88.73 J | =>

Now applying energy conservation between paint B&D" EB- Wfriction - Energy stored in spring 88-73 - fxxx Kv? 8873 - 6:97 xx = 1x7000 Kĩ Tummer advin. O EG = - 177.46 - 13.82 X = 7000 x 2 7000 x ² + 13.82x- 177.46 = 0 → 96 = -13.82 + (13.82)? - 4 x 7000 X (-177.46) S x=0.158 x=0.16 m 2X 7000 m I am ax²+bx+c=0 x=-bt N62_4 ac 2a ☺
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