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Physics 4 Review 1. 2. 3. 4. Name _________________________ Understand the concept of work and power. Work, W = F||d, measured in joules (W = Fcosd) When F and d , then + work When F and d , then – work When F and d , then 0 work (planet orbit) Work is a scalar (+ or – refers to amount) Work equals area under the F vs. d graph Power, P = W/t = Fvav, measured in watts Understand the concept of mechanical energy. Kinetic energy, K = ½mv2, measured in joules Gravitational potential energy, Ug = mgh, when close to the earth's surface or Ug = -GMm/r between planets Spring potential energy, Us = ½kx2 Contrast conservative and non-conservative forces. Conservative forces, such as gravity and spring convert potential energy kinetic energy o W = F||d, where d = displacement o When F and d , then U K When F and d , then K U Non-conservative forces, such as friction and push/pull change the total mechanical energy o W = F||d, where d = distance o When F and d , then increase Energy (+W) When F and d , then decrease Energy (-W) Work/Mechanical Energy model Spring System Us = ½kx2 9. 10. 11. Spring Force, Fs (Conservative force) Object K = ½mv2 Gravitational Force, Fg (Conservative force) 12. Gravity System Ug = mgh 5. 6. 7. 8. Push or Pull Force, Fp (Non-conservative force) Earth Solving conservation of mechanical energy problems. Problems with conservative forces only ½mv12 + mgh1 + ½kx12 = ½mv22 + mgh2 + ½kx22 problems with non-conservative forces K1 + U1 ± Wp – Wf = K2 + U2 ½mv12 + mgh1 + ½kx12 ± F||d = ½mv22 + mgh2 + ½kx22 Understand linear momentum and its relationship to force and kinetic energy. Linear momentum, p = mv, measured in kg•m/s Momentum is a vector (+ or – refers to direction) Impulse force, J = Ft = mv = p kinetic energy momentum is K = p2/2m Solve conservation of linear momentum problems. Momentum is conserved when there are no external forces involved Explosion problems: (mA + mB)v = mAvA' + mBvB' Inelastic collision, where objects stick after colliding: mAvA + mBvB = (mA + mB)v' Elastic collision (two formulas): vA + vA' = vB + vB' and mAvA + mBvB = mAvA' + mBvB' Solve collisions in two dimensions problems. mAvAx + mBvBx = (mA + mB)vx' or mAvAx' + mBvBx' mAvAy + mBvBy = (mA + mB)vy' or mAvAy' + mBvBy' 13. Solve an energy problem with conservation forces only. A spring (k = 1000 N/m) is compressed 0.10 m and is used to launch a 2.0-kg ball upward from a table top 1.0-m above the floor. The spring takes 0.15 s to launch the ball. a. What is the ball's velocity just after it is launched? b. How much power is generated by the spring? c. How high does the ball go? d. How fast is the ball traveling when it hits the floor? e. How fast would the ball be traveling when it hits the floor if it was initially launched horizontally? Solve an energy problem with non-conservative forces. The same spring (k = 1000 N/m) is used to launch a 5.0-kg block. The spring is compressed 0.20 m and released. The block first slides along a frictionless surface then across a rough section of flooring ( = 0.30). a. Graph Fs vs. distance for the decompressing spring. (1) What does the slope of the graph represent? (2) What does the area under the graph represent? b. How much work is done on the block by the spring (include sign in the answer)? c. What is the velocity of the block just after it is released from the spring? d. What is the force of friction on the block? e. How far does the block slide along the rough flooring before it comes to a halt? Solve an Inelastic collision problem. A 3-kg block slides on a frictionless table at 5 m/s and strikes a 5-kg block at rest on the edge of a table. The two blocks stick together and slide off of the table and strike the floor 1 m below. a. What is the total momentum of the two blocks before the collision? b. What is the velocity of the combined blocks after the collision? c. How much time are the two blocks in the air? d. How far from the base of the table do the two blocks land? Solve an semi-elastic collision problem. A 1-kg puck traveling due east at 3 m/s strikes a 3-kg pluck at rest. The 1-kg block slides at 2 m/s in the direction of 30o south of east. a. Complete the table. puck before the collision after the collision 1 kg 3 kg 1 kg 3 kg px py b. What is the resulting velocity of the 3-kg puck after the collision? c. What is the resulting direction of the 3-kg puck after the collision? d. What is the percentage of the original kinetic energy is conserved after the collision? Solve ballistics problems. bullet strikes a heavy block, which either swings to a maximum height or slides a distance along a surface inelastic collision between bullet (m) and block (M): mvbullet = (M + m)v' Conservation of energy, swing: ½(M + m)v'2 = (M + m)gh Conservation of energy, slide: ½(M + m)v'2 = (M + m)gd