Physics
Physics
3rd Edition
ISBN: 9781259233616
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 10, Problem 112P

(a)

To determine

The velocity of pan just after the landing of oranges.

(a)

Expert Solution
Check Mark

Answer to Problem 112P

The velocity is 2.18m/s.

Explanation of Solution

The spring constant is 450N/m, mass of pan is 0.250kg, mass of oranges is 2.20kg, and height of drop is 30.0cm.

Write the equation for law of conservation of energy for the fall of oranges.

12mvoi2=mgh

Here, m is the mass of oranges, voi is the speed of oranges at a moment just before it lands on the pan, g is the gravitational acceleration, and h is the height of fall.

Rewrite the above equation in terms of v.

voi=2gh (I)

Write the equation for law of conservation of momentum for before and after the landing of oranges on pan.

mvoi=(m+m)v (II)

Here, voi is the velocity of oranges before landing, m is the mass of pan, and v is the speed of pan with oranges.

Conclusion:

Substitute 9.80m/s2 for g and 30.0cm for h in equation (I) to find voi.

voi=2(9.80m/s2)(30.0cm(1m100cm))=2.4249m/s

Substitute 2.4249m/s for voi, 2.20kg for m, and 0.250kg for m in equation (II) to find v.

(2.20kg)(2.4249m/s)=(2.20kg+0.250kg)vv=5.33478kgm/s2.45kg=2.177m/s2.18m/s

Therefore, the velocity is 2.18m/s.

(b)

To determine

The new equilibrium point of pan with oranges from the position before the oranges was dropped.

(b)

Expert Solution
Check Mark

Answer to Problem 112P

The distance to new equilibrium point is 4.8cm.

Explanation of Solution

The spring constant is 450N/m, mass of pan is 0.250kg, mass of oranges is 2.20kg, and height of drop is 30.0cm.

Write the equilibrium condition for the old equilibrium point.

mg=kd

Here, k is the spring constant and d is the stretched distance.

Write the equation for new equilibrium on adding certain amount of mass.

(Δm)g=k(Δd)

Here, Δm is the additional mass and the new equilibrium distance.

Conclusion:

Substitute 2.20kg for Δm, 9.80m/s2 for g, and 450N/m for k in the above equation to find Δd.

(2.20kg)(9.80m/s2)=(450N/m)(Δd)Δd=21.56kgm/s2450N/m=0.0479m(100cm1m)4.8cm

Therefore, the distance to new equilibrium point is 4.8cm.

(c)

To determine

The amplitude of the oscillations.

(c)

Expert Solution
Check Mark

Answer to Problem 112P

The amplitude is 17cm.

Explanation of Solution

The spring constant is 450N/m, mass of pan is 0.250kg, mass of oranges is 2.20kg, and height of drop is 30.0cm.

Write the equation for law of conservation of energy.

Ki+Ui=Kf+Uf (I)

Here, Ki is the kinetic energy at the mean position of oscillation, Ui is the potential energy at the mean position.Kf is the kinetic energy at the extreme position, and Uf is the potential energy at the extreme position.

Write the equation for Ki.

Ki=12(m+m)v2 (II)

Write the equation for Ui.

Ui=12k(Δd)2 (III)

Write the equation for Uf.

Uf=12kA2 (IV)

Rewrite equation (I) by substituting equations (II), (III), (IV) and 0 for Kf.

12(m+m)v2+12k(Δd)2=0+12kA2

Eliminate the common terms and rewrite the above expression in terms of A.

A=(m+m)v2k+(Δd)2

Conclusion:

Substitute 2.20kg for m, 0.250kg for m, 2.17m/s for v, 450N/m for k, and 4.8cm for Δd in the above equation to find A.

A=(2.20kg+0.250kg)(2.17m/s)2450N/m+(4.8cm(102m1cm))2=0.0289m=0.17m(100cm1m)17cm

Therefore, the amplitude is 17cm.

(d)

To determine

The frequency of oscillations of the system.

(d)

Expert Solution
Check Mark

Answer to Problem 112P

The frequency is 2.2Hz.

Explanation of Solution

The spring constant is 450N/m, mass of pan is 0.250kg, mass of oranges is 2.20kg, and height of drop is 30.0cm.

Write the equation for the oscillations of a spring.

f=12πk(m+m)

Here, f is the frequency of oscillations of spring.

Conclusion:

Substitute 3.14 for π, 2.20kg for m, and 0.250kg for m, and 450N/m for k in the above equation to find Δd.

f=12(3.14)450N/m(2.20kg+0.250kg)=(0.159)(13.55N/mkg)=2.15Hz2.2Hz

Therefore, the frequency is 2.2Hz.

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