A spring has a length of 0.210 m when a 0.300-kg mass hangs from it, and a length of 0.800 m when a 1.95-kg mass hangs from it. What is the value of spring constant in N/m? [answer with three significant figures] [Assume g = 9.80 m/s2

Review. This problem extends the reasoning of Problem 41 in Chapter 9. Two gliders are set in motion on an air track. Glider 1 has mass m1 = 0.240 kg and moves to the right with speed 0.740 m/s. It will have a rear-end collision with glider 2, of mass m2 = 0.360 kg, which initially moves to the right with speed 0.120 m/s. A light spring of force constant 45.0 N/m is attached to the back end of glider 2 as shown in Figure P9.41. When glider 1 touches the spring, superglue instantly and permanently makes it stick to its end of the spring. (a) Find the common speed the two gliders have when the spring is at maximum compression. (b) Find the maximum spring compression distance. The motion after the gliders become attached consists of a combination of (1) the constant-velocity motion of the center of mass of the two-glider system found in part (a) and (2) simple harmonic motion of the gliders relative to the center of mass. (c) Find the energy of the center-of-mass motion. (d) Find the energy of the oscillation.

The position of a particle attached to a vertical spring is given by y=(y0cost)j. The y axis points upward, y0 = 14.5 cm. and = 18.85 rad/s. Find the position of the particle at a. t = 0 and b. t = 9.0 s. Give your answers in centimeters.

A spring 1.50 m long with force constant 475 N/m is hung from the ceiling of an elevator, and a block of mass 10.0 kg is attached to the bottom of the spring. (a) By how much is the spring stretched when the block is slowly lowered to its equilibrium point? (b) If the elevator subsequently accelerates upward at 2.00 m/s2, what is the position of the block, taking the equilibrium position found in part (a) as y = 0 and upwards as the positive y-direction. (c) If the elevator cable snaps during the acceleration, describe the subsequent motion of the block relative to the freely falling elevator. What is the amplitude of its motion?

A mass is placed on a frictionless, horizontal table. A spring (k=100N/m) , which can be stretched or compressed, is placed on the table. A 5.00-kg mass is attached to one end of the spring, the other end is anchored to the wall. The equilibrium position is marked at zero. A student moves the mass out to x=4.0 cm and releases it from rest. The mass oscillates in SHM. (a) Determine the equations of motion. (b) Find the position, velocity, and acceleration of the mass at time t=3.00 s.

A block of mass M rests on a table. It is fastened to the lower end of a light, vertical spring. The upper end of the spring is fastened to a block of mass m. The upper block is pushed down by an additional force 3mg, so the spring compression is 4mg/k. In this configuration, the upper block is released from rest. The spring lifts the lower block off the table. In terms of m, what is the greatest possible value for m?

A lightweight spring with spring constant k = 225 N/m is attached to a block of mass m1 = 4.50 kg on a frictionless, horizontal table. The blockspring system is initially in the equilibrium configuration. A second block of mass m2 = 3.00 kg is then pushed against the first block, compressing the spring by x = 15.0 cm as in Figure P16.77A. When the force on the second block is removed, the spring pushes both blocks to the right. The block m2 loses contact with the springblock 1 system when the blocks reach the equilibrium configuration of the spring (Fig. P16.77B). a. What is the subsequent speed of block 2? b. Compare the speed of block 1 when it again passes through the equilibrium position with the speed of block 2 found in part (a). 77. (a) The energy of the system initially is entirely potential energy. E0=U0=12kymax2=12(225N/m)(0.150m)2=2.53J At the equilibrium position, the total energy is the total kinetic energy of both blocks: 12(m1+m2)v2=12(4.50kg+3.00kg)v2=(3.75kg)v2=2.53J Therefore, the speed of each block is v=2.53J3.75kg=0.822m/s (b) Once the second block loses contact, the first block is moving at the speed found in part (a) at the equilibrium position. The energy 01 this spring-block 1 system is conserved, so when it returns to the equilibrium position, it will be traveling at the same speed in the opposite direction, or v=0.822m/s. FIGURE P16.77

You attach a block to the bottom end of a spring hanging vertically. You slowly let the block move down and find that it hangs at rest with the spring stretched by 15.0 cm. Next, you lift the block back up to the initial position and release it from rest with the spring unstretched. What maximum distance does it move down? (a) 7.5 cm (b) 15.0 cm (c) 30.0 cm (d) 60.0 cm (e) The distance cannot be determined without knowing the mass and spring constant.

A blockspring system oscillates with an amplitude of 3.50 cm. The spring constant is 250 N/m and the mass of the block is 0.500 kg. Determine (a) the mechanical energy of the system, (b) the maximum speed of the block, and (c) the maximum acceleration.

A 1.50-kg box rests atop a massless vertical spring with k = 4250 N/m that has been compressed by 15.0 cm from its equilibrium position. The box is released and leaves the spring when it reaches its equilibrium position. What is the maximum height the box reaches above its original position?