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

Question

EXAMPLE 13.4The Object-Spring System Revisited GOAL Apply the time-independent velocity expression,
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 xinto 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. SOLUTION Substitute x = A = 3.00 cm andk = 20.0 N/m into the equation for the total mechanical energy E. E = KE + PEg + PEs Use conservation of energy with xi = A and xf = 0 to compute the speed of the object at the origin. (KE + PEg + PEs)i = (KE + PEg + PEs)f 0 + 0 + ½kA2 = ½mv2max + 0 + 0 (B) Compute the velocity of the object when the displacement is 2.00 cm. Substitute known values. v = ± k/m(A2 x2) v = ± ( 20.0 N/m · [(0.0300 m)2 (0.0200 m )2] ) ½ 0.500 kg = ±0.141 m/s Substitute known values. v = ± k/m(A2 x2) v = ± ( 20.0 N/m · [(0.0300 m)2 (0.0200 m )2] ) ½ 0.500 kg = ±0.141 m/s Substitute known values. v = ± k/m(A2 x2) v = ± ( 20.0 N/m · [(0.0300 m)2 (0.0200 m )2] ) ½ 0.500 kg = ±0.141 m/s v = ± k/m(A2 x2) v = ± ( 20.0 N/m · [(0.0300 m)2 (0.0200 m )2] ) ½ 0.500 kg = ±0.141 m/s Substitute into the equation for kinetic energy. Substitute into the equation for spring potential energy. LEARN MORE PRACTICE IT Use the worked example above to help you solve this problem. A 0.465 kg object connected to a light spring with a spring constant of 22.0 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 PRACTICE IT Use the worked example above to help you solve this problem. A 0.465 kg object connected to a light spring with a spring constant of 22.0 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 EXERCISEHINTS:  GETTING STARTED  |  I'M STUCK! For what values of x is the speed of the object 0.15 m/s?
x = ± cm EXERCISEHINTS:  GETTING STARTED  |  I'M STUCK! For what values of x is the speed of the object 0.15 m/s?
x = ± cm k m (A2 - x2)

Explanation / Answer

(a) Ep = ½kx² = ½ * 22.0N/m * (0.03m)² = 0.0099 J = 9.9 mJ
Ek = 0.0099 J = ½mv² = ½ * 0.465kg * v²
v = 0.206 m/s

(b) v = ± sqrt[(k/m)*(A^2 – x^2)] = ± sqrt[(22/0.465)*(0.03^2 – 0.02^2)] = ±0.154 m/s

If the object is moving to the right, the sign is positive, ifit is moving to the left, the sign is negative.

(c) KE = 0.5mv^2 = 0.5*0.465*0.154^2 = 0.005514 J = 5.51 mJ

PE = 0.5kx^2 = 0.5*22*0.02^2 = 0.0044 J = 4.4 mJ

(d) 0.5mv^2 = 0.5kx^2

0.5*0.465*0.15^2 = 0.5*22*x^2

x = ±0.0218 m = ±2.18 cm

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