Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
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Chapter 7, Problem 69P

(a)

To determine

The distance traveled by block along the incline plane.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of block is 2.4kg .

The initial speed of block is 3.8m/s .

The coefficient of friction is 0.30 .

The angle of incline with the horizontal is 37° .

Formula used:

Write the expression for friction force along the incline.

  f=μmgcosθ

Here, m is mass of block, g is gravitational acceleration, μ is the coefficient of friction, f is the friction force and θ is the angle of weight of block with the perpendicular to the incline.

Write the expression for thermal energy.

  ΔEthermal=fΔs

Here, ΔEthermal is the change in thermal energy and Δs is the displacement covered by block.

Substitute μmgcosθ for f in above expression.

  ΔEthermal=(μmgcosθ)Δs

Write the expression for change in potential energy.

  ΔU=mg(hfhi)

Here, ΔU is the potential energy stored in the body, hi is the initial height of block and hf is the final height of block.

Write the expression for change in kinetic energy.

  ΔK=12m(vf2vi2)

Here, ΔK is the change in kinetic energy, vi is the initial velocity of block and vf is the final velocity of block.

Total energy of block is conserved at all points. Work done by external force is equal to the sum of change in gravitational potential energy, kinetic energy and thermal energy.

Write the expression of work done by external force.

  Wexternal=ΔU+ΔK+ΔEthermal ........(1)

Here, Wexternal is the work done by external force.

  Physics for Scientists and Engineers, Vol. 1, Chapter 7, Problem 69P , additional homework tip  1

There is no external force acting on the block; so, the work done by external force is zero.

Substitute 12m(vf2vi2) for ΔK , (μmgcosθ)Δs for ΔEthermal , 0 for Wexternal and mg(hfhi) for ΔU in equation (1).

  0=12m(vf2vi2)+mg(hfhi)+(μmgcosθ)Δs

Substitute Δssinθ for hf , 0 for hi , 0 for vf and rearrange the above expression in terms of Δs .

  Δs=vi22g(sinθ+μcosθ) ........(2)

Calculation:

Substitute 3.8m/s for vi , 0.30 for μ , 37° for θ and 9.81m/s2 for g in equation (2).

  Δs= ( 3.8m/s )22( 9.81m/ s 2 )( sin37°+( 0.30 )cos37°)=0.87m

Conclusion:

Thus, the distance traveled by block along the incline is 0.87m .

(b)

To determine

The speed of the block when it has traveled half the distance in part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of block is 2.4kg .

The initial speed of block is 3.8m/s .

The coefficient of friction is 0.30 .

The angle of incline with the horizontal is 37° .

Formula used:

Write the expression for friction force along the incline.

  f=μmgcosθ

Here, m is mass of block, g is gravitational acceleration, μ is the coefficient of friction, f is the friction force and θ is the angle of weight of block with the perpendicular to the incline.

Write the expression for thermal energy.

  ΔEthermal=fΔs

Here, ΔEthermal is the change in thermal energy and Δs is the displacement covered by block.

Substitute μmgcosθ for f in above expression.

  ΔEthermal=(μmgcosθ)Δs

Write the expression for change in potential energy.

  ΔU=mg(hfhi)

Here, ΔU is the potential energy stored in the body, hi is the initial height of block and hf is the final height of block.

Write the expression for change in kinetic energy.

  ΔK=12m(vf2vi2)

Here, ΔK is the change in kinetic energy, vi is the initial velocity of block and vf is the final velocity of block.

Total energy of block is conserved at all points. Work done by external force is equal to the sum of change in gravitational potential energy, kinetic energy and thermal energy.

Write the expression of work done by external force.

  Wexternal=ΔU+ΔK+ΔEthermal

Here, Wexternal is the work done by external force.

  Physics for Scientists and Engineers, Vol. 1, Chapter 7, Problem 69P , additional homework tip  2

There is no external force acting on the block; so, the work done by external force is zero.

Substitute 12m(vf2vi2) for ΔK , (μmgcosθ)Δs for ΔEthermal , 0 for Wexternal and mg(hfhi) for ΔU in equation (1).

  0=12m(vf2vi2)+mg(hfhi)+(μmgcosθ)Δs

Substitute Δssinθ for hf , 0 for hi , 0 for vf and rearrange the above expression in terms of Δs .

  Δs=vi22g(sinθ+μcosθ)

For object velocity at halfway on incline plane:

Substitute 12m(vf2vi2) for ΔK , (μmgcosθ)Δs2 for ΔEthermal , 0 for Wexternal and mg(hfhi) for ΔU in equation (1).

  0=12m(vf2vi2)+mg(hfhi)+(μmgcosθ)Δs2

Substitute (Δs2)sinθ for hf , 0 for hi and rearrange the above expression in terms of vf .

  vf=vi2gΔs(sinθ+μcosθ) ........(3)

Calculation:

Substitute 3.8m/s for vi , 0.30 for μ , 0.87m for Δs , 37° for θ and 9.81m/s2 for g in equation (3).

  vf= ( 3.8m/s )2( 9.81m/ s 2 )( 0.87m)[sin37°+( 0.30)cos37°]=2.7m/s

Conclusion:

Thus, the speed of the block when it has traveled half the distance in part (a)is 2.7m/s .

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Chapter 7 Solutions

Physics for Scientists and Engineers, Vol. 1

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