EXAMPLE 13.4 The Object-Spring System Revisited Goal Apply the time-independent

Question

EXAMPLE 13.4 The Object-Spring System Revisited Goal Apply the time-independent velocity expression,

to an object-spring system.

Problem A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. (b) What is the velocity of the object when the displacement is 2.00 cm? (c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm.

Strategy The total energy of the system can be found most easily at the amplitude x = A, where the kinetic energy is zero. There, the potential energy alone is equal to the total energy. Conservation of energy then yields the speed at x = 0. For part (b), obtain the velocity by substituting the given value of x into the time independent velocity equation. Using this result, the kinetic energy asked for in part (c) can be found by substitution, as can the potential energy.




Use the worked example above to help you solve this problem. A 0.505 kg object connected to a light spring with a spring constant of 21.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. E = J vmax = m/s
(b) What is the velocity of the object when the displacement is 2.00 cm?
m/s

(c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. KE = J PEs = J For what values of x is the speed of the object 0.12 m/s?
x = cm
EXAMPLE 13.4 The Object-Spring System Revisited Goal Apply the time-independent velocity expression,

to an object-spring system.

Problem A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. (b) What is the velocity of the object when the displacement is 2.00 cm? (c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm.

Strategy The total energy of the system can be found most easily at the amplitude x = A, where the kinetic energy is zero. There, the potential energy alone is equal to the total energy. Conservation of energy then yields the speed at x = 0. For part (b), obtain the velocity by substituting the given value of x into the time independent velocity equation. Using this result, the kinetic energy asked for in part (c) can be found by substitution, as can the potential energy.




Use the worked example above to help you solve this problem. A 0.505 kg object connected to a light spring with a spring constant of 21.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. E = J vmax = m/s
(b) What is the velocity of the object when the displacement is 2.00 cm?
m/s

(c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. KE = J PEs = J For what values of x is the speed of the object 0.12 m/s?
x = cm
EXAMPLE 13.4 The Object-Spring System Revisited Goal Apply the time-independent velocity expression,

to an object-spring system.

Problem A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. (b) What is the velocity of the object when the displacement is 2.00 cm? (c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm.

Strategy The total energy of the system can be found most easily at the amplitude x = A, where the kinetic energy is zero. There, the potential energy alone is equal to the total energy. Conservation of energy then yields the speed at x = 0. For part (b), obtain the velocity by substituting the given value of x into the time independent velocity equation. Using this result, the kinetic energy asked for in part (c) can be found by substitution, as can the potential energy.




Use the worked example above to help you solve this problem. A 0.505 kg object connected to a light spring with a spring constant of 21.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. E = J vmax = m/s
(b) What is the velocity of the object when the displacement is 2.00 cm?
m/s

(c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. KE = J PEs = J For what values of x is the speed of the object 0.12 m/s?
x = cm
Use the worked example above to help you solve this problem. A 0.505 kg object connected to a light spring with a spring constant of 21.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. E = J vmax = m/s
(b) What is the velocity of the object when the displacement is 2.00 cm?
m/s

(c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. KE = J PEs = J (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. E = J vmax = m/s
(b) What is the velocity of the object when the displacement is 2.00 cm?
m/s

(c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. KE = J PEs = J For what values of x is the speed of the object 0.12 m/s?
x = cm For what values of x is the speed of the object 0.12 m/s?
x = cm E = J vmax = m/s EXAMPLE 13.4 The Object-Spring System Revisited Goal Apply the time-independent velocity expression, to an object-spring system. Problem A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. (b) What is the velocity of the object when the displacement is 2.00 cm? (c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. Strategy The total energy of the system can be found most easily at the amplitude x = A, where the kinetic energy is zero. There, the potential energy alone is equal to the total energy. Conservation of energy then yields the speed at x = 0. For part (b), obtain the velocity by substituting the given value of x into the time independent velocity equation. Using this result, the kinetic energy asked for in part (c) can be found by substitution, as can the potential energy. Use the worked example above to help you solve this problem. A 0.505 kg object connected to a light spring with a spring constant of 21.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. E = J vmax = m/s (b) What is the velocity of the object when the displacement is 2.00 cm? m/s (c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. KE = J PEs = J For what values of x is the speed of the object 0.12 m/s? x = cm

Explanation / Answer

a)E=0.5*k*A^2=0.5*21*(3*10^-2)^2=9.45*10^-3j 0.5*k*A^2=0.5*m*v^2 v(max)=0.193 m/s b)v=+/- sqrt[(k/m)*(A^2-x^2)] =+/- 0.144 m/s c)ke=0.5*0.505*0.144^2=5.23810^-3 j pe=0.5*k*x^2=0.5*21*0.02^2=4.2*10^-3 j 0.12=+/- sqrt[(k/m)*(A^2-x^2)] 0.0144=(21/0.505)*(0.03^2-x^2) x=0.023 m=2.3 cm

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