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A 2.0-kg cart attached to a spring undergoes
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- Consider the data for a block of mass m = 0.250 kg given in Table P16.59. Friction is negligible. a. What is the mechanical energy of the blockspring system? b. Write expressions for the kinetic and potential energies as functions of time. c. Plot the kinetic energy, potential energy, and mechanical energy as functions of time on the same set of axes. Problems 5965 are grouped. 59. G Table P16.59 gives the position of a block connected to a horizontal spring at several times. Sketch a motion diagram for the block. Table P16.59arrow_forwardA 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 particle of mass m moving in one dimension has potential energy U(x) = U0[2(x/a)2 (x/a)4], where U0 and a are positive constants. (a) Find the force F(x), which acts on the particle. (b) Sketch U(x). Find the positions of stable and unstable equilibrium. (c) What is the angular frequency of oscillations about the point of stable equilibrium? (d) What is the minimum speed the particle must have at the origin to escape to infinity? (e) At t = 0 the particle is at the origin and its velocity is positive and equal in magnitude to the escape speed of part (d). Find x(t) and sketch the result.arrow_forward
- A 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_forwardThe 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.arrow_forwardA horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 47.0 J and a maximum displacement from equilibrium of 0.240 m. (a) What is the spring constant? (b) What is the kinetic energy of the system at the equilibrium point? (c) If the maximum speed of the block is 3.45 m/s, what is its mass? (d) What is the speed of the block when its displacement is 0.160 m? (e) Find the kinetic energy of the block at x = 0.160 m. (f) Find the potential energy stored in the spring when x = 0.160 m. (g) Suppose the same system is released from rest at x = 0.240 m on a rough surface so that it loses 14.0 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant?arrow_forward
- Review. A 0.250-kg block resting on a frictionless, horizontal surface is attached to a spring whose force constant is 83.8 N/m as in Figure P15.15. A horizontal force F causes the spring to stretch a distance of 5.46 cm from its equilibrium position. (a) Find the magnitude of F. (b) What is the total energy stored in the system when the spring is stretched? (c) Find the magnitude of the acceleration of the block just after the applied force is removed. (d) Find the speed of the block when it first reaches the equilibrium position. (e) If the surface is not frictionless but the block still reaches the equilibrium position, would your answer to part (d) be larger or smaller? (f) What other information would you need to know to find the actual answer to part (d) in this case? (g) What is the largest value of the coefficient of friction that would allow the block to reach the equilibrium position? Figure P15.15arrow_forwardA horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 47.0 J and a maximum displacement from equilibrium of 0.240 m. (a) What is the spring constant? (b) What is the kinetic energy of the system at the equilibrium point? (c) If the maximum speed of the block is 3.45 m/s, what is its mass? (d) What is the speed of the block when its displacement is 0.160 m? (e) Find the kinetic energy of the block at x = 0.160 m. (f) Find the potential energy stored in the spring when x = 0.160 m. (g) Suppose the same system is released from rest at x = 0.240 m on a rough surface so that it loses 14.0 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant?arrow_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_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 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.arrow_forwardA 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_forward
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