COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
2nd Edition
ISBN: 9781319414597
Author: Freedman
Publisher: MAC HIGHER
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Chapter 6, Problem 43QAP
To determine

(a)

The work done by the gravitational force on the statue.

Expert Solution
Check Mark

Answer to Problem 43QAP

The work done by the gravitational force on the statue is 2.83×103J.

Explanation of Solution

Given:

Mass of the crate

  m=150 kg

Angle made by the inclined plane with the horizontal

  θ=40.0°

Displacement of the crate along the plane

  d=3.0i^ m

Coefficient of kinetic friction between the crate and the plane μk=0.54

Formula used:

Draw a free body diagram representing the forces and apply the condition for dynamic equilibrium. Work done by a force is given by the product of the force and the displacement along the direction of force.

Calculation:

Draw the free body diagram for the forces and assume the positive direction of the x axis down the plane.

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 6, Problem 43QAP

Figure 1

The gravitational force FG acts vertically downwards, the normal force n acts upwards perpendicular to the plane along the + y axis and the curator applies the force FC up the plane, along the − x axis. Since the crate slides down the plane and makes a displacement d along the +x direction, the force of kinetic friction Ff acts up the plane along the − x axis.

The magnitude of the gravitational force is given by,

  FG=mg.......(1)

Resolve the gravitational force FG into two components FGx and FGy along +x and −y directions.

Therefore,

  FGx=FGsinθi^=mgsinθi^.....(2)

  FGy=FGcosθ(j^)=mgcosθj^.......(3)

The work done by the x component of the gravitational force is given by,

  WGx=FGx·d=(mgsinθi^)·d

Substitute the known values of the variables in the above equation.

  WGx=(mgsinθi^)·d=[(150 kg)(9.80  m/s 2)(sin40.0°)i^]·(3.0i^ m)=2.83×103 J

The work done by the y component of the gravitational force is given by,

  WGy=FGy·d=(mgcosθj^)·d

Substitute the known values of the variables in the above equation.

  WGy=(mgcosθj^)·d=[(150 kg)(9.80  m/s 2)(cos40.0°)j^]·(3.0i^ m)=0

Therefore the work done by the gravitational force is given by,

  WG=WGx+WGy=2.83×103 J+0 J=2.83×103 J

Conclusion:

Thus the work done by the gravitational force on the statue is 2.83×103J.

To determine

(b)

Work done by the Curator in pushing the statue up the incline.

Expert Solution
Check Mark

Answer to Problem 43QAP

The work done by the Curator in pushing the statue up the incline is 1.01×103J.

Explanation of Solution

Given:

Mass of the crate

  m=150 kg

Angle made by the inclined plane with the horizontal

  θ=40.0°

Displacement of the crate along the plane

  d=3.0i^ m

Coefficient of kinetic friction between the crate and the plane μk=0.54

Calculation:

The crate moves with a constant velocity, hence it is in dynamic equilibrium. The sum of the forces along the x and the y directions, independently add up to zero.

Use Fig 1, and apply the condition of equilibrium along the y axis.

  n+FGy=0

From equation (3) FGy=FGcosθ(j^)=mgcosθj^,

  n=FGy=(mgcosθj^)=mgcosθj^.....(4)

The magnitude of the force of friction and the normal force are related as follows:

  Ff=μkn

From equation (4),

  Ff=μkn=μkmgcosθ

The force of friction acts along the − x axis.

Therefore,

  Ff=μkmgcosθ(i^)=μkmgcosθi^.....(5)

Apply the condition of equilibrium along the x direction.

  FGx+FC+Ff=0

Therefore,

  FC=FGxFf

Use equations (2) FGx=FGsinθi^=mgsinθi^ and (5) Ff=μkmgcosθi^ in the above equation.

  FC=FGxFf=mgsinθi^(μkmgcosθi^)=mg(sinθμkcosθ)i^

Substitute the known values of the variables in the above equation.

  FC=mg(sinθμkcosθ)i^=(150 kg)(9.80  m/s2)[(sin40.0°)(0.54)(cos40.0°)]i^=3.368×102 Ni^

Write the expression for the work done by the Curator.

  WC=FC·d

Substitute the values of the variables in the above equation.

  WC=FC·d=(3.368× 102 Ni^)·(3.0i^ m)=1.01×103J

Conclusion:

Thus the work done by the Curator in pushing the statue up the incline is 1.01×103J.

To determine

(c)

The work done by the friction force on the crate

Expert Solution
Check Mark

Answer to Problem 43QAP

The work done by the friction force on the crate is 1.82×103J.

Explanation of Solution

Given:

Mass of the crate

  m=150 kg

Angle made by the inclined plane with the horizontal

  θ=40.0°

Displacement of the crate along the plane

  d=3.0i^ m

Coefficient of kinetic friction between the crate and the plane μk=0.54

Formula used:

The work done by the force of friction is given by,

  Wf=Ff·d

Calculation:

Use equation (5) Ff=μkmgcosθi^ in the formula.

  Wf=Ff·d=(μkmgcosθi^)·d

Substitute the known values of the variables in the equation.

  Wf=(μkmgcosθi^)·d=[(0.54)(150 kg)(9.80  m/s 2)(cos40.0°)i^]·(3.0i^ m)=1.82×103J

Conclusion:

Thus, the work done by the friction force on the crate is 1.82×103J.

To determine

(d)

The work done by the normal force between the crate and the incline.

Expert Solution
Check Mark

Answer to Problem 43QAP

The work done by the normal force between the crate and the incline is 0.

Explanation of Solution

Given:

The expressions for normal force and displacement.

  n=nj^d=di^

Formula used:

The work done by the normal force is given by,

  Wn=n·d

Calculation:

Substitute the given values of the vectors in the formula.

  Wn=n·d=nj^·di^=0

Conclusion:

Thus the work done by the normal force between the crate and the incline is 0.

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Chapter 6 Solutions

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

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