COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 4, Problem 92P

(a)

To determine

The acceleration of each blocks and tension in the cord.

(a)

Expert Solution
Check Mark

Answer to Problem 92P

The acceleration of each blocks is (m2gμkm1g)(m1+m2)_ and tension in the cord is m2gm2a_.

Explanation of Solution

COLLEGE PHYSICS-CONNECT ACCESS, Chapter 4, Problem 92P

Write the expression for the Newton’s second law for the block of mass m1.

    Tfk=m1a        (I)

Here, T is the tension in the cord, fk is the frictional force, m1 is the mass of block 1, a is the acceleration of each block.

Write the expression for the fk.

    fk=μkN=μkm1g        (II)

Here, μk is the coefficient of kinetic friction, g is the acceleration due to gravity.

Use equation (II) in (I) to solve for T.

    T=m1a+μkm1g        (III)

Write the expression for the Newton’s second law for the mass m2.

    m2gT=m2a        (IV)

Here, m2 is the mass of block 2.

Use equation (IV) to solve for T.

    T=m2gm2a        (V)

Equate equations (III) and (IV) to solve a.

    a=(m2gμkm1g)(m1+m2)        (VI)

Conclusion:

Therefore, the acceleration of each blocks is (m2gμkm1g)(m1+m2)_ and tension in the cord is m2gm2a_.

(b)

To determine

The acceleration and tension in the cord for m1m2, m1m2 and m1=m2.

(b)

Expert Solution
Check Mark

Answer to Problem 92P

The acceleration for m1m2 is g_, m1m2 is μkg_ and m1=m2 is g2(1μk)_ and the tension in the cord for m1m2 is 0_, m1m2 is m2g(1+μk)_ and m1=m2 is mg2(1+μk)_.

Explanation of Solution

Use equation (VI) to solve for acceleration for m1m2.

    a=(m2gμkm1g)(m1+m2)m2gm2=g        (VII)

Use equation (VI) to solve for acceleration for m1m2.

    a=(m2gμkm1g)(m1+m2)μkm1gm1=μkg        (VIII)

Use equation (VI) to solve for acceleration for m1=m2.

    a=(m2gμkm1g)(m1+m2)=mg(1μk)2m=g2(1μk)        (IX)

Use equation (V) to solve for the tension in the cord for m1m2.

    T=m2gm2a=m2gm2g=0        (X)

Use equation (V) to solve for the tension in the cord for m1m2.

    T=m2gm2a=m2gm2(μkg)=m2g(1+μk)        (XI)

Use equation (V) to solve for the tension in the cord for m1=m2.

    T=m2gm2a=m2gm2[g2(1μk)]=m2g2+μkm2g2=mg2(1+μk)        (XII)

Conclusion:

Therefore, the acceleration for m1m2 is g_, m1m2 is μkg_ and m1=m2 is g2(1μk)_ and the tension in the cord for m1m2 is 0_, m1m2 is m2g(1+μk)_ and m1=m2 is mg2(1+μk)_.

(c)

To determine

The tension in the two blocks if the blocks slides at a constant velocity.

(c)

Expert Solution
Check Mark

Answer to Problem 92P

The tension in the two blocks if the blocks slides at a constant velocity is m2g_.

Explanation of Solution

Use equation (V) to solve for the tension in the cord for the block sliding at a constant velocity.

    T=m2gm2(0)=m2g        (XIII)

Conclusion:

Therefore, the tension in the two blocks if the blocks slides at a constant velocity is m2g_.

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Chapter 4 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

Ch. 4.4 - Prob. 4.8PPCh. 4.5 - Prob. 4.5ACPCh. 4.5 - Prob. 4.10PPCh. 4.5 - Prob. 4.5BCPCh. 4.6 - Prob. 4.11PPCh. 4.6 - Prob. 4.6CPCh. 4 - Prob. 1CQCh. 4 - Prob. 2CQCh. 4 - Prob. 3CQCh. 4 - Prob. 4CQCh. 4 - Prob. 5CQCh. 4 - Prob. 6CQCh. 4 - Prob. 7CQCh. 4 - Prob. 8CQCh. 4 - Prob. 9CQCh. 4 - Prob. 10CQCh. 4 - Prob. 11CQCh. 4 - Prob. 12CQCh. 4 - Prob. 13CQCh. 4 - Prob. 14CQCh. 4 - Prob. 15CQCh. 4 - Prob. 16CQCh. 4 - Prob. 17CQCh. 4 - Prob. 18CQCh. 4 - Prob. 19CQCh. 4 - Prob. 1MCQCh. 4 - Prob. 2MCQCh. 4 - Prob. 3MCQCh. 4 - Prob. 4MCQCh. 4 - Prob. 5MCQCh. 4 - Prob. 6MCQCh. 4 - Prob. 7MCQCh. 4 - Prob. 8MCQCh. 4 - Prob. 9MCQCh. 4 - Prob. 10MCQCh. 4 - Prob. 11MCQCh. 4 - Prob. 12MCQCh. 4 - Prob. 13MCQCh. 4 - Prob. 14MCQCh. 4 - Prob. 15MCQCh. 4 - Prob. 16MCQCh. 4 - Prob. 17MCQCh. 4 - Prob. 18MCQCh. 4 - Prob. 19MCQCh. 4 - Prob. 20MCQCh. 4 - Prob. 21MCQCh. 4 - Prob. 1PCh. 4 - Prob. 4PCh. 4 - Prob. 5PCh. 4 - Prob. 6PCh. 4 - Prob. 7PCh. 4 - Prob. 8PCh. 4 - Prob. 9PCh. 4 - Prob. 10PCh. 4 - Prob. 11PCh. 4 - Prob. 12PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19PCh. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 61PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69PCh. 4 - Prob. 70PCh. 4 - Prob. 71PCh. 4 - Prob. 72PCh. 4 - Prob. 73PCh. 4 - Prob. 74PCh. 4 - Prob. 75PCh. 4 - Prob. 76PCh. 4 - Prob. 77PCh. 4 - Prob. 78PCh. 4 - Prob. 79PCh. 4 - Prob. 80PCh. 4 - Prob. 81PCh. 4 - Prob. 82PCh. 4 - Prob. 83PCh. 4 - Prob. 84PCh. 4 - Prob. 85PCh. 4 - Prob. 86PCh. 4 - Prob. 87PCh. 4 - Prob. 88PCh. 4 - Prob. 90PCh. 4 - Prob. 91PCh. 4 - Prob. 92PCh. 4 - Prob. 93PCh. 4 - Prob. 94PCh. 4 - Prob. 95PCh. 4 - Prob. 97PCh. 4 - Prob. 98PCh. 4 - Prob. 99PCh. 4 - Prob. 100PCh. 4 - Prob. 101PCh. 4 - Prob. 102PCh. 4 - Prob. 103PCh. 4 - Prob. 104PCh. 4 - Prob. 105P
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