Chapter 6, Problem 43QAP

To determine

(a)

The work done by the gravitational force on the statue.

Answer to Problem 43QAP

The work done by the gravitational force on the statue is $2.83\times {10}^3\text{J}$.

Explanation of Solution

Given:

Mass of the crate

  $m=150\text{ kg}$

Angle made by the inclined plane with the horizontal

  $\theta =40.0°$

Displacement of the crate along the plane

  $\overrightarrow{d}=3.0\widehat{i}\text{ m}$

Coefficient of kinetic friction between the crate and the plane ${\mu }_k=0.54$

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.

Figure 1

The gravitational force ${\overrightarrow{F}}_G$ acts vertically downwards, the normal force $\overrightarrow{n}$ acts upwards perpendicular to the plane along the + y axis and the curator applies the force ${\overrightarrow{F}}_C$ up the plane, along the − x axis. Since the crate slides down the plane and makes a displacement $\overrightarrow{d}$ along the +x direction, the force of kinetic friction ${\overrightarrow{F}}_f$ acts up the plane along the − x axis.

The magnitude of the gravitational force is given by,

  $F_G=mg$.......(1)

Resolve the gravitational force ${\overrightarrow{F}}_G$ into two components ${\overrightarrow{F}}_{Gx}$ and ${\overrightarrow{F}}_{Gy}$ along +x and −y directions.

Therefore,

  ${\overrightarrow{F}}_{Gx}=F_G\sin \theta \widehat{i}=mg\sin \theta \widehat{i}$.....(2)

  ${\overrightarrow{F}}_{Gy}=F_G\cos \theta \left(-\widehat{j}\right)=-mg\cos \theta \widehat{j}$.......(3)

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

  $\begin{array}{c}{W}_{Gx}={\overrightarrow{F}}_{Gx}·\overrightarrow{d}\\ =\left(mg\sin \theta \widehat{i}\right)·\overrightarrow{d}\end{array}$

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

  $\begin{array}{c}{W}_{Gx}=\left(mg\sin \theta \widehat{i}\right)·\overrightarrow{d}\\ =\left[\left(150\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)\left(\sin 40.0°\right)\widehat{i}\right]·\left(3.0\widehat{i}\text{ m}\right)\\ =2.83\times {10}^3\text{ J}\end{array}$

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

  $\begin{array}{c}{W}_{Gy}={\overrightarrow{F}}_{Gy}·\overrightarrow{d}\\ =\left(-mg\cos \theta \widehat{j}\right)·\overrightarrow{d}\end{array}$

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

  $\begin{array}{c}{W}_{Gy}=\left(-mg\cos \theta \widehat{j}\right)·\overrightarrow{d}\\ =-\left[\left(150\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)\left(\cos 40.0°\right)\widehat{j}\right]·\left(3.0\widehat{i}\text{ m}\right)\\ =0\end{array}$

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

  $\begin{array}{c}{W}_G=W_{Gx}+W_{Gy}\\ =2.83\times {10}^3\text{ J}+0\text{ J}\\ \text{=}2.83\times {10}^3\text{ J}\end{array}$

Conclusion:

Thus the work done by the gravitational force on the statue is $2.83\times {10}^3\text{J}$.

To determine

(b)

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

Answer to Problem 43QAP

The work done by the Curator in pushing the statue up the incline is $-1.01\times {10}^3\text{J}$.

Explanation of Solution

Given:

Mass of the crate

  $m=150\text{ kg}$

Angle made by the inclined plane with the horizontal

  $\theta =40.0°$

Displacement of the crate along the plane

  $\overrightarrow{d}=3.0\widehat{i}\text{ m}$

Coefficient of kinetic friction between the crate and the plane ${\mu }_k=0.54$

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.

  $\overrightarrow{n}+{\overrightarrow{F}}_{Gy}=0$

From equation (3) ${\overrightarrow{F}}_{Gy}=F_G\cos \theta \left(-\widehat{j}\right)=-mg\cos \theta \widehat{j}$,

  $\begin{array}{c}\overrightarrow{n}=-{\overrightarrow{F}}_{Gy}\\ =-\left(-mg\cos \theta \widehat{j}\right)\\ =mg\cos \theta \widehat{j}\end{array}$.....(4)

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

  $F_f={\mu }_{k}n$

From equation (4),

  $\begin{array}{c}{F}_f={\mu }_{k}n\\ ={\mu }_{k}mg\cos \theta \end{array}$

The force of friction acts along the − x axis.

Therefore,

  ${\overrightarrow{F}}_f={\mu }_{k}mg\cos \theta \left(-\widehat{i}\right)=-{\mu }_{k}mg\cos \theta \widehat{i}$.....(5)

Apply the condition of equilibrium along the x direction.

  ${\overrightarrow{F}}_{Gx}+{\overrightarrow{F}}_C+{\overrightarrow{F}}_f=0$

Therefore,

  ${\overrightarrow{F}}_C=-{\overrightarrow{F}}_{Gx}-{\overrightarrow{F}}_f$

Use equations (2) ${\overrightarrow{F}}_{Gx}=F_G\sin \theta \widehat{i}=mg\sin \theta \widehat{i}$ and (5) ${\overrightarrow{F}}_f=-{\mu }_{k}mg\cos \theta \widehat{i}$ in the above equation.

  $\begin{array}{c}{\overrightarrow{F}}_C=-{\overrightarrow{F}}_{Gx}-{\overrightarrow{F}}_f\\ =-mg\sin \theta \widehat{i}-\left(-{\mu }_{k}mg\cos \theta \widehat{i}\right)\\ =-mg\left(\sin \theta -{\mu }_k\cos \theta \right)\widehat{i}\end{array}$

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

  $\begin{array}{c}{\overrightarrow{F}}_C=-mg\left(\sin \theta -{\mu }_k\cos \theta \right)\widehat{i}\\ =-\left(150\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)\left[\left(\sin 40.0°\right)-\left(0.54\right)\left(\cos 40.0°\right)\right]\widehat{i}\\ =-3.368\times {10}^2\text{ N}\widehat{i}\end{array}$

Write the expression for the work done by the Curator.

  $W_C={\overrightarrow{F}}_C·\overrightarrow{d}$

Substitute the values of the variables in the above equation.

  $\begin{array}{c}{W}_C={\overrightarrow{F}}_C·\overrightarrow{d}\\ =\left(-3.368\times {10}^2\text{ N}\widehat{i}\right)·\left(3.0\widehat{i}\text{ m}\right)\\ =-1.01\times {10}^3\text{J}\end{array}$

Conclusion:

Thus the work done by the Curator in pushing the statue up the incline is $-1.01\times {10}^3\text{J}$.

To determine

(c)

The work done by the friction force on the crate

Answer to Problem 43QAP

The work done by the friction force on the crate is $-1.82\times {10}^3\text{J}$.

Explanation of Solution

Given:

Mass of the crate

  $m=150\text{ kg}$

Angle made by the inclined plane with the horizontal

  $\theta =40.0°$

Displacement of the crate along the plane

  $\overrightarrow{d}=3.0\widehat{i}\text{ m}$

Coefficient of kinetic friction between the crate and the plane ${\mu }_k=0.54$

Formula used:

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

  $W_f={\overrightarrow{F}}_f·\overrightarrow{d}$

Calculation:

Use equation (5) ${\overrightarrow{F}}_f=-{\mu }_{k}mg\cos \theta \widehat{i}$ in the formula.

  $\begin{array}{c}{W}_f={\overrightarrow{F}}_f·\overrightarrow{d}\\ =\left(-{\mu }_{k}mg\cos \theta \widehat{i}\right)·\overrightarrow{d}\end{array}$

Substitute the known values of the variables in the equation.

  $\begin{array}{c}{W}_f=\left(-{\mu }_{k}mg\cos \theta \widehat{i}\right)·\overrightarrow{d}\\ =\left[-\left(0.54\right)\left(150\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)\left(\cos 40.0°\right)\widehat{i}\right]·\left(3.0\widehat{i}\text{ m}\right)\\ =-1.82\times {10}^3\text{J}\end{array}$

Conclusion:

Thus, the work done by the friction force on the crate is $-1.82\times {10}^3\text{J}$.

To determine

(d)

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

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.

  $\begin{array}{c}\overrightarrow{n}=n\widehat{j}\\ \overrightarrow{d}=d\widehat{i}\end{array}$

Formula used:

The work done by the normal force is given by,

  $W_n=\overrightarrow{n}·\overrightarrow{d}$

Calculation:

Substitute the given values of the vectors in the formula.

  $\begin{array}{c}{W}_n=\overrightarrow{n}·\overrightarrow{d}\\ =n\widehat{j}·d\widehat{i}\\ =0\end{array}$

Conclusion:

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