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Jf 15 J of potential energy are stored in a spring which has been compressed by 2 cm from its relaxed length, how much force will this spring exert if it is stretched by 4 cm from its relaxed length? A) 500 N B) 3000 N C) 7500 N D) 75,000 N E) None of the Above

Jf 15 J of potential energy are stored in a spring which has been compressed by 2 cm from its relaxed length, how much force will this spring exert if it is stretched by 4 cm from its relaxed length? A) 500 N B) 3000 N C) 7500 N D) 75,000 N E) None of the Above

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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7. A 2-kg object moving to the right with velocity 5 m/s collides with a 3-kg object which is initially at rest. After the collision, the 3-kg object has a velocity vector y = i-j. What is the speed of the 2-kg object after the collision? A) 0.8 m/s B) 1.8 m/s C) 2.8 m/s D) 3.8 m/s E) 4.8 m/s 8. If 15 J of potential energy are stored in a spring which has been compressed by 2 cm from its relaxed length, how much force will this spring exert if it is stretched by 4 cm from its relaxed length? A) 500 N B) 3000 N C) 7500 N D) 75,000 N E) None of the Above mass of 10 kg is attached to the spring in problem 8. The spring is compressed by 2 cm and is released. When the spring reaches its equilibrium position, what the speed of the 10-kg mass? A) 1 m/s B) 1.7 m/s C) 2 m/s D) 2.7 m/s E) 3 m/s 3.S + 1.S S 10. A toy car is on a small indoor track, which is 100-m in circumference. The car starts from rest, accelerates uniformly, and after 5 seconds reaches a point halfway around the circle. acceleration? this time, what is the magnitude of the car's A) 4 m/s B) 9 m/s C) 16 m/s D) 25 m/s E) >25 m/s 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 12.25+ 2.25 14.5
Transcribed Image Text:7. A 2-kg object moving to the right with velocity 5 m/s collides with a 3-kg object which is initially at rest. After the collision, the 3-kg object has a velocity vector y = i-j. What is the speed of the 2-kg object after the collision? A) 0.8 m/s B) 1.8 m/s C) 2.8 m/s D) 3.8 m/s E) 4.8 m/s 8. If 15 J of potential energy are stored in a spring which has been compressed by 2 cm from its relaxed length, how much force will this spring exert if it is stretched by 4 cm from its relaxed length? A) 500 N B) 3000 N C) 7500 N D) 75,000 N E) None of the Above mass of 10 kg is attached to the spring in problem 8. The spring is compressed by 2 cm and is released. When the spring reaches its equilibrium position, what the speed of the 10-kg mass? A) 1 m/s B) 1.7 m/s C) 2 m/s D) 2.7 m/s E) 3 m/s 3.S + 1.S S 10. A toy car is on a small indoor track, which is 100-m in circumference. The car starts from rest, accelerates uniformly, and after 5 seconds reaches a point halfway around the circle. acceleration? this time, what is the magnitude of the car's A) 4 m/s B) 9 m/s C) 16 m/s D) 25 m/s E) >25 m/s 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 12.25+ 2.25 14.5
v² Circular motion: a = R W Pavr = Δt Fnet = ma; Weight: Fg = mg, ; g = 9.8 m/s?; 1 Kinetic energy: K = -m v²; 2 Potential energy: Ug=mgy E = Er E = U+K v2 Rotational motion: 0 = 1 rev = 2n rad; v = o r; a, = = w?r; at = a r; r 0 = 0o + at; 0 = 0o t+ 2θα-ω - 0 2 T=rxF; T= rFsino; Στ-Ια; Iy=ml 1 Ipoinț mass = mr? Iaisk = mR 2 1 Irod(end) = -mL? 2 Ishell = =mR? 3 Ihoop = mR? Irod(center) = mL? Iball = =mR? 12 work: W=r0; K=-lo 2 dW P = dt I= Icom + MD² W = Pavr = At 1 m Vcom ? Rolling: 2 K = 2 Vcom = Ro T= f,R F,max = H,Fn Incline: Fg=mgsin0 F mgcos0 Angular momentum: Lpoint mass = m rxv L= mrysin 0: L=m (r,vy - ryVx)k L= Io m¡X1+m2X2 m, +m2 LI = Lr I 01 =I 202 X com miy1+m2y2 Y com = m1 +m2 12.257
Transcribed Image Text:v² Circular motion: a = R W Pavr = Δt Fnet = ma; Weight: Fg = mg, ; g = 9.8 m/s?; 1 Kinetic energy: K = -m v²; 2 Potential energy: Ug=mgy E = Er E = U+K v2 Rotational motion: 0 = 1 rev = 2n rad; v = o r; a, = = w?r; at = a r; r 0 = 0o + at; 0 = 0o t+ 2θα-ω - 0 2 T=rxF; T= rFsino; Στ-Ια; Iy=ml 1 Ipoinț mass = mr? Iaisk = mR 2 1 Irod(end) = -mL? 2 Ishell = =mR? 3 Ihoop = mR? Irod(center) = mL? Iball = =mR? 12 work: W=r0; K=-lo 2 dW P = dt I= Icom + MD² W = Pavr = At 1 m Vcom ? Rolling: 2 K = 2 Vcom = Ro T= f,R F,max = H,Fn Incline: Fg=mgsin0 F mgcos0 Angular momentum: Lpoint mass = m rxv L= mrysin 0: L=m (r,vy - ryVx)k L= Io m¡X1+m2X2 m, +m2 LI = Lr I 01 =I 202 X com miy1+m2y2 Y com = m1 +m2 12.257
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