Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 10, Problem 95P

(a)

To determine

The spring constant of the spring.

(a)

Expert Solution
Check Mark

Answer to Problem 95P

The spring constant of the spring is 98.0N/m.

Explanation of Solution

The length of the spring is 20.0cm, the mass of the brick hanged is 1.10kg, the stretch of the spring is 31.0cm.

Write the expression for total force in vertical y-direction on the spring.

Fy=0kdmg=0k=mgd

Here, k is the spring constant, m is the mass hanged, d is the stretched distance of the spring at equilibrium, g is acceleration due to gravity.

Conclusion:

Substitute 1.10kg for m, 9.80m/s2 for g and 31.0cm20.0cm for d to find the spring constant of the spring.

k=(1.10kg)(9.80m/s2)31.0cm20.0cm=(1.10kg)(9.80m/s2)11.0cm102m1.0cm=98.0N/m

(b)

To determine

The maximum speed of the brick.

(b)

Expert Solution
Check Mark

Answer to Problem 95P

The maximum speed of the brick is 0.472m/s.

Explanation of Solution

The length of the spring is 20.0cm, the mass of the brick hanged is 1.10kg, the stretch of the spring is 31.0cm, the pulled length of brick from equilibrium point is 5.00cm.

Write the expression for conservation of energy to find the maximum speed of the brick.

Kmax=Umax12mvmax2=12kd2vmax=kd2m

Here, Kmax is the maximum kinetic energy, Umax is the maximum potential energy, vmax is the maximum speed of the brick.

Conclusion:

Substitute 1.10kg for m, 98.0N/m for k and 5.00cm for d to find the maximum speed of the brick.

vmax=(98.0N/m)(5.00cm102m1.0cm)21.10kg=0.472m/s

(c)

To determine

The speed of the brick when the displacement is 2.50cm.

(c)

Expert Solution
Check Mark

Answer to Problem 95P

The speed of the brick when the displacement is 2.50cm is 0.409m/s.

Explanation of Solution

The length of the spring is 20.0cm, the mass of the brick hanged is 1.10kg, the stretch of the spring is 31.0cm.

Write the expression for conservation of energy to find the speed of the brick.

E=K+Uθ12kd2=12mv2+12ky2v=km(d2y2)

Here, y is the displacement of the brick.

Conclusion:

Substitute 1.10kg for m, 98.0N/m for k, 5.00cm for d and 2.50cm for y to find the speed of the brick.

v=98.0N/m1.10kg((5.00cm102m1.0cm)2(2.50cm102m1.0cm)2)=0.409m/s

(d)

To determine

The time taken for the brick to oscillate five times.

(d)

Expert Solution
Check Mark

Answer to Problem 95P

The time taken for the brick to oscillate five times is 3.33s.

Explanation of Solution

The length of the spring is 20.0cm, the mass of the brick hanged is 1.10kg, the stretch of the spring is 31.0cm.

Write the expression for time taken for the brick to oscillate five times.

5T=5(2πmk)

Conclusion:

Substitute 1.10kg for m, 98.0N/m for k to find the time taken for the brick to oscillate five times.

5T=5(2π1.10kg98.0N/m)=10π1.10kg98.0N/m=3.33s

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Chapter 10 Solutions

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