Constant Acceleration Equations (1) x(t) = x + vt + at² (2) v(t)=v₂ + at (3) v² = +2a(x-x₂) W=F,Ar +F,Ar, (constant force) W=F|AF|cos (constant force) p=mv System Momentum Conservation m₂v₁ + m₂v₂ =m₂v₁ +₂³₂ Inelastic Collision V = V/2/ Elastic Collision V2/V₁=₁₁-21 ΣF.=ma, ΣF, =ma, ΣΕ =ma, f₁ = μ₂n Energy Equations my+mgy: [_my² + mgy₁ + = kx² + [W₁ = _ _ m² + 10² +mgyou + [W = 10² +mgy cst 2 K = -mv² U₂ = mgy -koc no slip condition: v= cor 12
Static Equilibrium @= T= r=t|F||F|sin 0 T=xF, -yF T>0=CCW r<0=CW ΣΕ = 0 ΣF, =0 Στ=0 Universal Gravity Gm,m, Simple Harmonic Motion m T= 2n₁ k 2πT T V max = 0 A = 00² A E max F=KA=ma max =mr² point particle L = 10 TH Ly Lynt c (hyp) A a (adj) ax² +bx+c=0 sin = opp/hyp=b/c b (opp) cos 0 adj/hyp= a/c tan = opp/adj = b/a c²=a² + b² Quadratic Equation X= -b± √b² - 4ac 2a
7. A string connects a block with mass m to a pulley, with mass m and radius r. A Hooke's Law spring with spring constant k=- connects the block to the floor. 5 mg 2 r See figure. Initially, the pulley has a CW angular momentum of -Lo, the spring is unstretched, and the block is moving upward with unknown speed vo. If the block turns around when it has moved up a distance r, find 1)r and 2) the magnitude of the tension force in the string. Model the pulley as a solid disk rotating about an axis through its center. Assume the string does not slip on the pulley. The knowns are m. g, and Lo. (20 points) Use two energy equations and FBDs to solve this problem. ceiling string = ? m floor m, r -Lo 4= ?
Constant Acceleration Equations (1) x(t) = x + vt + at² (2) v(t)=v₂ + at (3) v² = +2a(x-x₂) W=F,Ar +F,Ar, (constant forc
-
answerhappygod
- Site Admin
- Posts: 899604
- Joined: Mon Aug 02, 2021 8:13 am
Constant Acceleration Equations (1) x(t) = x + vt + at² (2) v(t)=v₂ + at (3) v² = +2a(x-x₂) W=F,Ar +F,Ar, (constant forc
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!