Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 8, Problem 59P

(a)

To determine

The maximum velocity of each block if the coefficient of kinetic friction between block and the surface is 0.

(a)

Expert Solution
Check Mark

Answer to Problem 59P

The maximum velocity of each block is 0.256i^m/s and 0.128i^m/s respectively.

Explanation of Solution

Given info: The force constant is 3.85N/m , the spring is compressed by 8.00cm . The mass of left and right side block is 0.250kg and 0.500kg respectively.

Write the expression to calculate the force by the spring.

Fs=kx (1)

Here,

Fs is the spring force.

k is spring constant.

x is the compression or expansion length.

Substitute 3.85N/m for k and 8.00cm for x in equation (1) to find Fs .

Fs=3.85N/m×8.00cm×1m100cm=0.308N

Write the expression of conservation of energy.

Espring=KLB+KRB12kx2=12m1v1f2+12m1v2f2 (2)

Here,

m1 is the mass of left block.

m2 is the mass of right block.

v1f is the velocity of the left block.

v2f is the velocity of the right block.

Substitute 3.85N/m for k , 8.00cm for x , 0.250kg for m1 and 0.500kg for m2 in equation (2).

12(3.85N/m)×(8.00cm×1m100cm)2=12(0.250kg)v1f2+12(0.500kg)v2f20.0123J=12(0.250kg)v1f2+12(0.500kg)v2f2 (3)

Write the expression of conservation of linear momentum.

0=m1v1f+(m2v2f) (4)

Substitute 0.250kg for m1 and 0.500kg for m2 in equation (4).

0=0.250kg×v1f+(0.500kg×v2f)v1f=2v2f (5)

Substitute 2v2f for v1f in equation (3) to find v2f .

0.0123J=12(0.250kg)(2v2f)2+12(0.500kg)v2f20.0123J=12(1.50kg)v2f2v2f=0.128i^m/s

Substitute 0.128i^m/s for v2f in equation (5) to find v1f .

v1f=2(0.128i^m/s)=0.256i^m/s

Thus, the maximum velocity of each block is 0.256i^m/s and 0.128i^m/s respectively.

Conclusion:

Therefore, the maximum velocity of each block is 0.256i^m/s and 0.128i^m/s respectively.

(b)

To determine

The maximum velocity of each block if the coefficient of kinetic friction between block and the surface is 0.100 .

(b)

Expert Solution
Check Mark

Answer to Problem 59P

The maximum velocity of each block is 0.0642i^m/s and 0 respectively.

Explanation of Solution

Given info: The force constant is 3.85N/m , the spring is compressed by 8.00cm . The mass of left and right side block is 0.250kg and 0.500kg respectively.

Write the expression to calculate the normal force on the lighter block.

R=m1a (6)

Here,

a is the acceleration of the left block.

Substitute 0.250kg for m1 and 9.8m/s2 for g in equation (6) to find R .

R=0.250kg×9.8m/s2=2.45N

Write the expression to calculate the limiting frictional force.

fk=μR

Here,

μ is the coefficient of kinetic friction.

Substitute 2.45N for R and 0.100 for μ in above equation.

fk=0.100×2.45N=0.245N

The spring force, 0.308N is greater than the limiting frictional force 0.245N . Therefore, the left block will move.

Since the mass of right block is double than the left block therefore the limiting force of friction is also two times the left one i.e. 2fk=0.490N .

The limiting frictional force of right clock is greater than the spring force so it will not move.

The left will continue to move as long as the spring force is large than the friction force.

Write the expression to calculate the limiting frictional force.

fk=kxf

Here,

xf is the maximum distance travelled by the left block.

Substitute 0.245N for fk and 3.85N/m for k in above equation.

0.245N=3.85N/m×xfxf=0.0636m

Write the expression of conservation of energy.

12kx2fk(xxf)=12m1v1f2+12kxf2 (7)

Substitute 3.85N/m for k , 8.00cm for x , 0.250kg for m1 , 0.245N for fk , 0.0636m for xf in equation (7) to find v1f .

[12(3.85N/m)×(8.00cm×1m100cm)2(0.245N)(8.00cm×1m100cm0.0636m)]=[12(0.250kg)v1f2+12(3.85N/m)(0.0636m)2]0.0123J-0.004J=12(0.250kg)v1f2+0.0078Jv1f=0.0642m/s=0.0642i^m/s

The negative sign indicates that the direction of motion of the lighter block is toward negative x axis.

The velocity of the heavier block is zero.

Thus, the maximum velocity of each block is 0.0642i^m/s and 0 respectively.

Conclusion:

Therefore, the maximum velocity of each block is 0.0642i^m/s and 0 respectively.

(c)

To determine

The maximum velocity of each block if the coefficient of kinetic friction between block and the surface is 0.462 .

(c)

Expert Solution
Check Mark

Answer to Problem 59P

The velocity of both blocks is 0.

Explanation of Solution

Given info: The force constant is 3.85N/m , the spring is compressed by 8.00cm . The mass of left and right side block is 0.250kg and 0.500kg respectively.

Write the expression to calculate the limiting frictional force of left block.

fk=μR=μm1a

Substitute 0.250kg for m1 and 9.8m/s2 for g and 0.462 for μ in above equation.

fk=0.462×0.250kg×9.8m/s2=1.13N

Write the expression to calculate the limiting frictional force of right block.

fk=μR=μm2a

Substitute 0.500kg for m2 and 9.8m/s2 for g and 0.462 for μ in above equation.

fk=0.462×0.500kg×9.8m/s2=2.26N

The spring force is less than the limiting frictional force of both the blocks so the blocks will not move.

Conclusion:

Therefore, the velocity of both blocks is 0.

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Chapter 8 Solutions

Principles of Physics: A Calculus-Based Text

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