COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 4, Problem 104P

(a)

To determine

The acceleration of the crate.

(a)

Expert Solution
Check Mark

Answer to Problem 104P

The acceleration of the crate is 0.31m/s2_.

Explanation of Solution

Let 25.0kg be m1, and 14.0kg be m2.

Write the expression for the frictional force acting on mass m2.

  f2=μm2g        (I)

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

Write the expression for frictional force acting on mass m1.

  f1=μ(m1gFsinθ)        (II)

Here, F is the applied force, θ is the angle inclined with mass m1.

Write the expression for the total force acting on m2.

  Tf2=m2a        (III)

Here, T is the tension.

Substitute, equation (I) in (III), and rearrange to obtain an expression for T.

  Tμm2g2=m2aT=m2(μg+a)        (IV)

Write the expression for total force acting on mass m1.

  FcosθTμ(m1gFsinθ)=m1a        (V)

Substitute equation (IV) in (V), and rearrange to obtain an expression for a.

  Fcosθm2(μg+a)μ(m1gFsinθ)=m1aa=F(cosθμsinθ)μg(m1+m2)m1+m2        (VI)

Conclusion:

Substitute, 195N for F, 20° for θ, 0.55 for μ, 25.0kg for m1, and 14.0kg for m2, and 9.8m/s2 for g in equation (VI).

  a=195N(cos20°+μsin20°)(0.55)(9.8m/s2)(25.0kg+14.0kg)(25.0kg+14.0kg)=0.31m/s2

Therefore, the acceleration of the crate is 0.31m/s2_.

(b)

To determine

The tension of the rope connecting two wires.

(b)

Expert Solution
Check Mark

Answer to Problem 104P

The tension of the rope connecting two wires is 80N_.

Explanation of Solution

Use equation (IV) to find the tension.

Conclusion:

Substitute, 14.0kg for m2, 0.55 for μ, 0.31m/s2 for a, and 9.8m/s2 for g in equation (IV) to find the tension.

  T=14.0kg(0.55(9.8m/s2)+0.31m/s2)=80N

Therefore, the tension of the rope connecting two wires is 80N_.

(c)

To determine

The distance moved by the crates in 3.0s.

(c)

Expert Solution
Check Mark

Answer to Problem 104P

The distance moved by the crates in 3.0s is 1.4m_.

Explanation of Solution

Write the kinematic equation for distance moved.

  s=υ0t+12at2        (VII)

Here, υ0 is the initial velocity.

Conclusion:

Substitute, 0 for υ0, 0.31m/s2 for a , and 3.0s for t in equation (VII).

  s=0+12(0.31m/s2)(3.0s)2=1.4m

Therefore, the distance moved by the crates in 3.0s is 1.4m_.

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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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