Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 5, Problem 65P

(a)

To determine

To show coefficient of kinetic friction μk=2Δx( Δt)2g .

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The time measure is Δt .

Total displacement of the block is Δx .

Formula used:

Draw free body diagram of block.

  Physics for Scientists and Engineers, Chapter 5, Problem 65P , additional homework tip  1

Write expression for net force in the x -direction.

  Fx=0

Substitute fkmax for Fx in above expression.

  fkmax=0

Here, fk is the friction for due to kinetic friction, m is the mass of the block and ax is the acceleration of the block.

Rearrange above expression for fk .

  fk=max

Substitute μkmg for fk in above expression and solve for ax .

  μkmg=maxax=μkg

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

Calculation:

Write expression for displacement of the block.

  Δx=vavΔt

Here, vav is the average velocity of the block and Δt is the time taken by the block.

Substitute (v1+v22) for vav in above expression.

  Δx=(v1+v22)Δt

Here, v1 is the initial speed of the block and v2 is the final speed of the block.

Substitute 0 for v2 in above expression and solve for v1Δt .

  Δx=( v 1 +02)Δtv1Δt=2Δx

Write expression for the displacement of the block.

  Δx=v1Δt+12ax(Δt)2

Substitute 2Δx for v1Δt in above expression.

  Δx=2Δx+12ax(Δt)2Δx=12ax(Δt)2

Substitute μkg for ax in above expression and solve for μk .

  Δx=12(μkg)(Δt)2μk=2Δxg ( Δt )2

Conclusion:

Thus, the coefficient of kinetic friction is μk=2Δx( Δt)2g .

(b)

To determine

The value of coefficient of kinetic friction.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The block slides 1.37m .

The time taken is 0.97s .

Formula used:

Draw free body diagram of block.

  Physics for Scientists and Engineers, Chapter 5, Problem 65P , additional homework tip  2

Write expression for net force in the x -direction.

  Fx=0

Substitute fkmax for Fx in above expression.

  fkmax=0

Rearrange above expression for fk .

  fk=max

Substitute μkmg for fk in above expression and solve for ax .

  μkmg=maxax=μkg

Write expression for displacement of the block.

  Δx=vavΔt

Substitute (v1+v22) for vav in above expression.

  Δx=(v1+v22)Δt

Substitute 0 for v2 in above expression and solve for v1Δt .

  Δx=( v 1 +02)Δtv1Δt=2Δx

Write expression for the displacement of the block.

  Δx=v1Δt+12ax(Δt)2

Substitute 2Δx for v1Δt in above expression.

  Δx=2Δx+12ax(Δt)2Δx=12ax(Δt)2

Substitute μkg for ax in above expression and solve for μk .

  μk=2Δxg( Δt)2.......(1)

Calculation:

Substitute 1.37m for Δx , 9.81m/s2 for g and 0.97s for Δt in equation (1).

  μk=2( 1.37m)( 9.81m/ s 2 ) ( 0.97s )2=0.30

Conclusion:

Thus, the value of the kinetic friction is 0.30 .

(c)

To determine

The initial speed of the block

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

Formula used:

Draw free body diagram of block.

  Physics for Scientists and Engineers, Chapter 5, Problem 65P , additional homework tip  3

Write expression for net force in the x -direction.

  Fx=0

Substitute fkmax for Fx in above expression.

  fkmax=0

Rearrange above expression for fk .

  fk=max

Substitute μkmg for fk in above expression and solve for ax .

  μkmg=maxax=μkg

Write expression for displacement of the block.

  Δx=vavΔt

Substitute (v1+v22) for vav in above expression.

  Δx=(v1+v22)Δt

Substitute 0 for v2 in above expression and solve for v1 .

  v1=2ΔxΔt.......(1)

Calculation:

Substitute 1.37m for Δx , and 0.97s for Δt in equation (1).

  v1=2( 1.37m)0.97s=2.8m/s

Conclusion:

Thus, the initial speed of the block is 2.8m/s .

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Chapter 5 Solutions

Physics for Scientists and Engineers

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