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- A grandfather clock has a pendulum length of 0.7 m and mass bob of 0.4 kg. A mass of 2 kg falls 0.8 m in seven days to keep the amplitude (from equilibrium) of the pendulum oscillation steady at 0.03 rad. What is the Q of the system?arrow_forwardA vibration sensor, used in testing a washing machine, consists of a cube of aluminum 1.50 cm on edge mounted on one end of a strip of spring steel (like a hacksaw blade) that lies in a vertical plane. The strips mass is small compared with that of the cube, but the strips length is large compared with the size of the cube. The other end of the strip is clamped to the frame of the washing machine that is not operating. A horizontal force of 1.43 N applied to the cube is required to hold it 2.75 cm away from its equilibrium position. If it is released, what is its frequency of vibration?arrow_forwardA block of mass m = 2.00 kg is attached to a spring of force constant k = 500 N/m as shown in Figure P7.15. The block is pulled to a position xi = 5.00 cm to the right of equilibrium and released from rest. Find the speed the block has as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface is k = 0.350. Figure P7.15arrow_forward
- 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 P16.77arrow_forwardUse the data in Table P16.59 for a block of mass m = 0.250 kg and assume friction is negligible. a. Write an expression for the force FH exerted by the spring on the block. b. Sketch FH versus t.arrow_forwardA 50.0-g object connected to a spring with a force constant of 35.0 N/m oscillates with an amplitude of 4.00 cm on a frictionless, horizontal surface. Find (a) the total energy of the system and (b) the speed of the object when its position is 1.00 cm. Find (c) the kinetic energy and (d) the potential energy when its position is 3.00 cm.arrow_forward
- An object of mass m1 = 9.00 kg is in equilibrium when connected to a light spring of constant k = 100 N/m that is fastened to a wall as shown in Figure P12.67a. A second object, m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.200 m (see Fig. P12.67b). The system is then released, and both objects start moving to the right on the frictionless surface. (a) When m1 reaches the equilibrium point, m2 loses contact with m1 (see Fig. P12.67c) and moves to the right with speed v. Determine the value of v. (b) How far apart are the objects when the spring is fully stretched for the first time (the distance D in Fig. P12.67d)? Figure P12.67arrow_forwardFor each expression, identify the angular frequency , period T, initial phase and amplitude ymax of the oscillation. All values are in SI units. a. y(t) = 0.75 cos (14.5t) b. vy (t) = 0.75 sin (14.5t + /2) c. ay (t) = 14.5 cos (0.75t + /2) 16.3arrow_forwardIn an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x=5.00cos(2t+6) where x is in centimeters and t is in seconds. At t = 0, find (a) the position of the piston, (b) its velocity, and (c) its acceleration. Find (d) the period and (e) the amplitude of the motion.arrow_forward
- The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?arrow_forwardA uniform wire (Y = 2.0 1011 N/m2) is subjected to a longitudinal tensile stress of 4.0 107 N/m2. What is the fractional change in the length of the wire?arrow_forwardWhich of the following statements is not true regarding a massspring system that moves with simple harmonic motion in the absence of friction? (a) The total energy of the system remains constant. (b) The energy of the system is continually transformed between kinetic and potential energy. (c) The total energy of the system is proportional to the square of the amplitude. (d) The potential energy stored in the system is greatest when the mass passes through the equilibrium position. (e) The velocity of the oscillating mass has its maximum value when the mass passes through the equilibrium position.arrow_forward
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