Q23E - University Physics with Modern Physics

Question

You are a member of an Alpine rescue team. You must project a box of supplies up an incline of constant slope angle  so that it reaches a standard skier who is a vertical distance h above the bottom of the incline. The incline is slippery, but there is some friction present with kinetic friction coefficient  ${\mathbf{\mu }}_{\mathbf{k}}$ . Use the the work energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of g, h${\mathbf{\mu }}_{\mathbf{k}}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{\alpha }$.

Step-by-Step Solution

Answer

Minimum speed with which the  box should be sent upward is $v_{1A}=\sqrt{2g\left({\mu }_{k}s \cos \alpha +h\right)}$

Step 1: Identification of the given data

The given data is listed below as-

• Coefficient of kinetic friction is ${\mu }_k$ 

Step 2: Significance of the work energy theorem

When forces act on a particle it undergoes displacement and the work done by these forces is equal to the change in kinetic energy of the particle. Therefore, work done on the particle is given by-

${\mathbf{W}}_{\mathbf{total}}\mathbf{=}{\mathbf{K}}_{\mathbf{2}}\mathbf{-}{\mathbf{K}}_{\mathbf{1}}\mathbf{=}\mathbf{∆}\mathbf{K}$ 

Here, $\mathbf{∆}\mathbf{K}$  is the change in kinetic energy of the body.

Step 3: Determination of the minimum speed that must be given to the box

The total work done on a particle is given by

$W_{total}=K_{2B}-K_{1A}=∆K....................\left(1\right)$ 

Where,  $K_{1A}=\frac{1}{2}m{v^2}_{1A}$ and  $K_{2B}=\frac{1}{2}m{v^2}_{2B}$ are the initial and final kinetic energy of the particle respectively.

Then, equation (1) becomes

$W_{total}=\frac{1}{2}m{v^2}_{2B}-\frac{1}{2}m{v^2}_{1A}$ 

Here $v_{2B}=0$ , since the minimum velocity is required for the box to reach upto a standard skier.

Therefore,

 

$$\begin{gathered} W_{total}=\frac{1}{2}m{v^2}_{2B}-\frac{1}{2}m{v^2}_{1A} \\ {W}_{total}=0-\frac{1}{2}m{v^2}_{1A} \\ {W}_{total}=0-\frac{1}{2}m{v^2}_{1A}.............................\left(2\right) \end{gathered}$$ 

Work done by kinetic force is given by 

 $W_f=f_k·s·\cos \varphi$ 

The force is in opposite direction to the motion, so angle   

 $$\begin{gathered} W_f=f_{k}scos({88}^o \\ =-f_{k}s \end{gathered}$$

Now, work done by the gravitational force will be 

$\begin{array}{l}{W}_g={\int }_{y_1}^{y_{2x}}{F}_g·ds\\ W_g={\int }_{y_1}^{y_2}{F}_g ds \cos \varphi \end{array}$ 

Here,  $y_1=0$ is the bottom of the incline plane and $y_2=h$ is the height of the skier and angle  $\varphi =180°$

 

$$\begin{gathered} W_g={\int }_0^h F_g ds \cos 180° \\ =-F_{g}h \\ =-mgh.........................\left(3\right) \end{gathered}$$ 

Now, total work done on the box is equal to work done by kinetic friction and gravitation.

$W_{tot}=W_f+W_g$  

Substitute value from equation (2) and (3) in above equation

$-\frac{1}{2}m{v^2}_{1A}=-\left({\mu }_{k}mg \cos \alpha \right)s-mgh$ 

Rearrange the above equation to obtain:

${v^2}_{1A}=2\left({\mu }_{k}gs \cos \alpha +gh\right)$  

Or,  $v_{1A}=\sqrt{2\left({\mu }_{k}gs \cos \alpha +gh\right)}$

$v_{1A}=\sqrt{2\left({\mu }_{k}s \cos \alpha +h\right)}$ 

 

Thus, the minimum speed that must be given to the box at the bottom of the incline so that it will reach the skier is $v_{1A}=\sqrt{2\left({\mu }_{k}s \cos \alpha +h\right)}$,