Question

A constant horizontal force moves a 50kgtrunk 6.0 m up a $\mathbf{30}\mathbf{°}$incline at constant speed. The coefficient of kinetic friction is 0.20. What are (a) the work done by the applied force and (b) the increase in the thermal energy of the trunk and incline?

Short answer

1. Work done by the applied horizontal force will be$2.2\times {10}^3\mathrm{J}$
2. An increase in thermal energy of the trunk and incline will be$7.7\times {10}^2\mathrm{J}$

Step-by-step solution

The given data

Mass of the trunk is,$\mathrm{m}=50\mathrm{kg}$

Distance covered by the trunk at the incline is,$\mathrm{x}=6.0\mathrm{m}$

The angle of inclination is,$\mathrm{\theta }=30°$

The coefficient of friction is,$\mathrm{\mu }=0.20$

Understanding the concept of kinematics and friction

We draw the free body diagram for the trunk. Using this, we can find the net force acting on the trunk. As we know the displacement of the trunk and the force are calculated, and we can find the work done by the applied force. Using the work done on a system by external force we can find the increase in the thermal energy.

Formulae:

The frictional force acting on a body, ${\mathrm{F}}_{\mathrm{f}}={\mathrm{\mu }}_{\mathrm{s}}{\mathrm{F}}_{\mathrm{N}}$ (1)

The normal force acting on a body, ${\mathrm{F}}_{\mathrm{N}}=\mathrm{mg}$ (2)

The work done by an applied force, $\mathrm{W}=\mathrm{Force}\times \mathrm{displacement}$ (3)

The force due to Newton’s second law, $\mathrm{F}=\mathrm{ma}$ (4)

a) Calculation of the work done by the horizontal force

Free body diagram,

From the free body diagram, we can say that,

${\mathrm{F}}_{\mathrm{N}}={\mathrm{F}}_1\mathrm{sin\theta }+\mathrm{mgcos\theta }..................\left(5\right)$

And

${\mathrm{F}}_1\mathrm{cos\theta }-{\mathrm{F}}_{\mathrm{f}}-\mathrm{mgsin\theta }=\mathrm{ma}$

As the trunk is moving with constant velocity, a = 0 thus, the above equation using equations (1), (2), and (4) becomes,

$$\begin{gathered} {\mathrm{F}}_1\mathrm{cos\theta }={\mathrm{F}}_{\mathrm{f}}+\mathrm{mgsin\theta } \\ {\mathrm{F}}_1\mathrm{cos\theta }=\left({\mathrm{\mu }}_{\mathrm{s}}+{\mathrm{F}}_{\mathrm{N}}\right)+\mathrm{mgsin\theta } \\ {\mathrm{F}}_1\mathrm{cos\theta }-\mathrm{mgsin\theta }=\left({\mathrm{\mu }}_{\mathrm{s}}+{\mathrm{F}}_{\mathrm{N}}\right) \\ \frac{\left({\mathrm{F}}_1\mathrm{cos\theta }-\mathrm{mgsin\theta }\right)}{{\mathrm{\mu }}_{\mathrm{s}}}={\mathrm{F}}_{\mathrm{N}} \\ {\mathrm{F}}_{\mathrm{N}}=\frac{\left({\mathrm{F}}_1\mathrm{cos\theta }-\mathrm{mgsin\theta }\right)}{{\mathrm{\mu }}_{\mathrm{s}}} \end{gathered}$$

But, using equation (5), the above equation giving the horizontal force can be written as:

role="math" localid="1661486810675" $$\begin{gathered} \frac{\left({\mathrm{F}}_1\mathrm{cos\theta }-\mathrm{mgsin\theta }\right)}{{\mathrm{\mu }}_{\mathrm{s}}}={\mathrm{F}}_1\mathrm{sin\theta }+\mathrm{mgcos\theta } \\ \left({\mathrm{F}}_1\mathrm{cos\theta }-\mathrm{mgsin\theta }\right)={\mathrm{\mu }}_{\mathrm{s}}\times {\mathrm{F}}_1\mathrm{sin\theta }+\mathrm{mgcos\theta } \\ {\mathrm{F}}_1\mathrm{cos\theta }-{\mathrm{\mu }}_{\mathrm{s}}{\mathrm{F}}_1\mathrm{sin\theta }={\mathrm{mg\mu }}_{\mathrm{s}}\mathrm{cos\theta }+\mathrm{mgsin\theta } \\ {\mathrm{F}}_1\times \left(\mathrm{cos\theta }-{\mathrm{\mu }}_{\mathrm{s}}\mathrm{sin\theta }\right)=\left({\mathrm{mg\mu }}_{\mathrm{s}}\mathrm{cos\theta }+\mathrm{mgsin\theta }\right) \\ {\mathrm{F}}_1=\frac{\left({\mathrm{mg\mu }}_{\mathrm{s}}\mathrm{cos\theta }+\mathrm{mgsin\theta }\right)}{\left(\mathrm{cos\theta }-{\mathrm{\mu }}_{\mathrm{s}}\mathrm{sin\theta }\right)} \\ {\mathrm{F}}_1=\frac{\left(50\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 0.20\times \cos 30°+50\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times \sin 30°\right)}{\left(\cos 30°-0.2\times \sin 30°\right)} \\ {\mathrm{F}}_1=434.1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right) \\ {\mathrm{F}}_1=434.1\mathrm{N} \end{gathered}$$

Now, the work done by this horizontal force using equation (3) is given as:

$$\begin{gathered} \mathrm{W}=\left(434.1\mathrm{N}\times 6\mathrm{m}\right)\cos 30° \\ =2.2\times {10}^3\mathrm{N}·\mathrm{m}\left(\frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}\right) \\ =2.2\times {10}^3\mathrm{J} \end{gathered}$$

Hence, the value of the work done is $2.2\times {10}^3\mathrm{J}$

b) Calculation of increase in the thermal energy

Change in the P.E. of the trunk from the tree diagram is given as:

$$\begin{gathered} ∆\mathrm{P}.\mathrm{E}.=\mathrm{mglsin\theta } \\ =\left(50\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(6\mathrm{m}\right)\sin 30° \\ =1.47\times {10}^3\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2\left(\frac{1\mathrm{J}}{1\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =1.47\times {10}^3\mathrm{J} \end{gathered}$$

Work done on a system by the external force is given by,

$\mathrm{W}=∆\mathrm{K}.\mathrm{E}.+∆\mathrm{P}.\mathrm{E}.+∆{\mathrm{E}}_{\mathrm{th}}$

In this case, the thermal energy can be calculated by substituting the values as:

$$\begin{gathered} \mathrm{W}=0+∆\mathrm{P}.\mathrm{E}.+∆{\mathrm{E}}_{\mathrm{th}} \\ ∆{\mathrm{E}}_{\mathrm{th}}=\mathrm{W}-∆\mathrm{P}.\mathrm{E}. \\ ∆{\mathrm{E}}_{\mathrm{th}}=2.24\times {10}^3\mathrm{J}-1.47\times {10}^3\mathrm{J} \\ ∆{\mathrm{E}}_{\mathrm{th}}=7.7\times {10}^2\mathrm{J} \end{gathered}$$

Hence, the value of the energy is $7.7\times {10}^2\mathrm{J}$