PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 6, Problem 65P

(a)

To determine

The spring constant.

(a)

Expert Solution
Check Mark

Answer to Problem 65P

Spring constant of the spring is 1.9N/cm.

Explanation of Solution

Write the expression for the force acting on the spring.

F=kx

Here, F is the force acting on the spring, k is the spring constant and x is the elongation or compression for the spring.

Write the expression for the weight hanging on the spring.

W=mg

Here, W is the weight of the mass, m is the mass that is hanging in equilibrium from the spring and g is the acceleration due to gravity.

Since weight is hanging from the string. Force acting on the spring is equal to weight of the mass.

Therefore, write the expression for force acting on the spring.

mg=kx

Rearrange above equation to get k.

k=mgx                                                                                                                          (I)

Conclusion:

Substitute 1.4kg for m, 9.80N/kg for g and 7.2cm for x in above equation to get k.

k=(1.4kg)(9.80N/kg)(7.2cm)=1.9N/cm

Therefore, spring constant of the spring is 1.9N/cm.

(b)

To determine

The elastic potential energy stored in the spring.

(b)

Expert Solution
Check Mark

Answer to Problem 65P

The elastic potential energy stored in the spring is 0.49J.

Explanation of Solution

The elastic potential energy stored in the spring.

Uelastic=12kx2

Here, Uelastic is the elastic potential energy stored in the spring.

Substitute mgx for k in above equation to get Uelastic.

Uelastic=12(mgx)x2=12mgx

Conclusion:

Substitute 1.4kg for m , 9.80N/kg for g and 7.2cm for x in above equation to get Uelastic..
Uelastic=12(1.4kg)(9.80N/kg)(7.2cm×1m100cm)=0.49J

Therefore, the elastic potential energy stored in the spring is 0.49J.

(c)

To determine

The second mass if it can produce an elongation of 12.2cm.

(c)

Expert Solution
Check Mark

Answer to Problem 65P

The second mass if it can produce an elongation of 12.2cm is 2.4kg.

Explanation of Solution

Let m be the new mass that hanging from spring and x is the elongation due to new mass m.

Write the expression for the spring constant in terms of new mass and elongation.

k=mgx (II)

Spring constant of the spring is a constant independent of force.

Conclusion:

Divide equation (II) by (I) to get m.

kk=mgxmgxm=mxx (III)

Substitute 1.4kg for m , 12.2cm for x2 and 7.2cm for x1 in above equation to get m.

m=(1.4kg)(12.2cm)(7.2cm)=2.4kg

Therefore, the second mass if it can produce an elongation of 12.2cm is 2.4kg.

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Chapter 6 Solutions

PHYSICS

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