11 please easy way to solve and use this formula sheet please v2Circular motion: a =RFnet = ma;Weight: Fg = mg, ;g = 9.8 m/s?;Payr =At1(Kinetic energy: K = –m v²;2Potential energy: Ug=mgyE = ErE = U+Kv2Rotational motion:1 rev = 2n rad;v = o r;= 0?r;at = a r;a, =1K=-lo2@ = 00 + at;0 = 0o t+22θα -ω -ω=rxFT= rFsino;1Ipoint massIgF mRΣτ-Ια;= mr?Idisk =mR+%3D22112Iball = =mR?2Ishell =3 MR2Ihoop = mR?Irod(center) =mL?Irod(end) =12mL?work: W=r0; K=-lo ?dWP =dtWPavr =AtI= Icom + MD?W =22Rolling:K= lo221m VcomVcom = RoT= f,RF,max = HFnIncline: F=mgsin0 F=mgcos0Angular momentum:Lpoint mass =m rxvL=mrvsin 0;L=m (r,Vy - ryVz)kL= IoLI= LI 01 =I 202m,x1 +m2X2miy1+m2y2X comY comm1 +m2m1 +m21 B) 9 m/sC) 16 m/sD) 25 m/sE) >25 m/s11. A 2-kg block starts from rest at the top of a 10-m long ramp inclined at 45 deg to the horizontal. The block is releasedand begins to slide down the ramp. At the bottom of the ramp the block has a speed of 9 m/s. What is the coefficient offriction between the block and the ramp?A) 0.12B) 0.22C) 0.32D) 0.42E) 0.52

11. A 2-kg block starts from rest at the top of a 10-m long ramp inclined at 45 deg to the horizontal. The block is released and begins to slide down the ramp. At the bottom of the ramp the block has a speed of 9 m/s. What is the coefficient of friction between the block and the ramp? A) 0.12 B) 0.22 C) 0.32 D) 0.42 E) 0.52 4

11. A 2-kg block starts from rest at the top of a 10-m long ramp inclined at 45 deg to the horizontal. The block is released and begins to slide down the ramp. At the bottom of the ramp the block has a speed of 9 m/s. What is the coefficient of friction between the block and the ramp? A) 0.12 B) 0.22 C) 0.32 D) 0.42 E) 0.52 4

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter8: Conservation Of Energy

Section: Chapter Questions

Problem 77PQ: A block of mass m1 = 4.00 kg initially at rest on top of a frictionless, horizontal table is...

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