The Physics of Everyday Phenomena
The Physics of Everyday Phenomena
8th Edition
ISBN: 9780073513904
Author: W. Thomas Griffith, Juliet Brosing Professor
Publisher: McGraw-Hill Education
Question
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Chapter 6, Problem 4SP

(a)

To determine

The initial potential energy.

(a)

Expert Solution
Check Mark

Answer to Problem 4SP

The initial potential energy is 3.5J.

Explanation of Solution

Given info: The spring constant is 360N/m and the stretched distance is 0.14m.

Write the expression for the elastic potential energy of a spring.

PE=12kx2 (1)

Here,

PE is the elastic potential energy

k is the spring constant

x is the stretched distance

Substitute 360N/m for k and 0.14m for x in equation (1) to find PE.

PE=12(360N/m)(0.14m)2=3.5J

Conclusion:

Thus the initial potential energy of the spring-mass system is 3.5J.

(b)

To determine

The maximum velocity that the mass will reach in its oscillation and the position where then system achieves maximum velocity.

(b)

Expert Solution
Check Mark

Answer to Problem 4SP

The maximum velocity that the mass will reach in its oscillation is 5.9m/s and this velocity occurs as the mass moves through the equilibrium position.

Explanation of Solution

Given info: The mass attached to the spring is 0.20kg.

The maximum velocity will be at a point where the kinetic energy of the system is maximum. The maximum kinetic energy is obtained when all the potential energy stored in the state, is completely converted to the kinetic energy. Since the initial potential energy is 3.5J, the maximum kinetic energy that the system can achieve is also 3.5J.

KE=3.5J

Write the expression for kinetic energy of an object.

KE=12mv2

Here,

m is the mass of the object

v is the velocity of the object

Rewrite the above equation for v.

v=2(KE)m (2)

Substitute 3.5J for KE and 0.20kg for m in equation (2) to find the value of maximum v.

v=2(3.5J)0.30kg=5.9m/s

The potential energy of the spring-mass system converts completely to kinetic energy when the system moves through the equilibrium position. This implies the maximum velocity occurs as the mass moves through the equilibrium position.

Conclusion:

Therefore the maximum velocity that the mass will reach in its oscillation is 5.9m/s and this velocity occurs as the mass moves through the equilibrium position.

(c)

To determine

The potential energy, kinetic energy and the velocity of the mass when the mass is 7cm from the equilibrium position.

(c)

Expert Solution
Check Mark

Answer to Problem 4SP

When the mass is 7cm from the equilibrium position, the potential energy is 0.88J, kinetic energy is 2.6J and the velocity is 5.1m/s.

Explanation of Solution

Given info: The mass attached to spring is 0.20kg and the spring constant is 360N/m.

Substitute 360N/m for k and 7cm for x in equation (1) to find PE.

PE=12(360N/m)(6cm(102m1cm))2=12(360N/m)(0.06 m)2=0.88J

At equilibrium position total energy is equal to the kinetic energy of the system. But as the system moves from the equilibrium position, it gains potential energy and there will be decrease in kinetic energy since total energy is constant. Hence the kinetic energy at a particular point is obtained by subtracting the potential energy at that point from the maximum kinetic energy.

Thus, the kinetic energy at the given point is obtained as,

KE=3.5J0.88J=2.6J

Substitute 2.6J for KE and 0.20kg for m in equation (2) to find the velocity at the given position.

v=2(2.6J)0.20kg=5.1m/s

Conclusion:

Thus when the mass is 7cm from the equilibrium position, the potential energy is 0.88J, kinetic energy is 2.6J and the velocity is 5.1m/s.

(d)

To determine

The comparison of velocity of the mass at the equilibrium position to the value when it is 7cm away from the equilibrium position.

(d)

Expert Solution
Check Mark

Answer to Problem 4SP

The velocity of the mass when it is 7cm away from the equilibrium position is 86 % of the maximum velocity.

Explanation of Solution

Take the ratio of the values of velocity of the mass when it is 7cm away from the equilibrium position to when it is at equilibrium position.

ratio=v7 cmveq

Here,

v7 cm is the velocity of the mass when it is 7cm away from the equilibrium position

veq is the velocity of the mass when it is at the equilibrium position

Substitute 5.1m/s for v7 cm and 5.9m/s for veq in the above equation to determine the ratio.

v7 cmveq=5.1m/s5.9m/s=0.86v7 cm=0.86veq

Thus, the velocity of the mass at the position 7cm away from the equilibrium position is 86% of that at equilibrium position.

Conclusion:

Thus the velocity of the mass when it is 7cm away from the equilibrium position is 86 % of the maximum velocity.

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