Q65P - University Physics with Modern Physics

Question

Consider the blocks in Exercise 6.7 as they move 75 cm. Find the total work done on each one (a) if there is no friction between the table and the 20.0 N block, and (b) if ${\mathit{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{500}$and ${\mathit{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{325}$between the table and the 20.0-N block.

                                      

Step-by-Step Solution

Answer

1. the work done by the first crate and second crate is 5.66 J and 3.4 J respectively, when friction is present.
2. the work done by the first crate and second crate is 2.57 J and 1.54 J respectively, when friction is present.

Step 1: Identification of given data

The given data can be listed below,

• The distance traveled by the block is, $d=75 \mathrm{cm}$
• The mass of the first crate, sliding on the table is, $m_1=20 \mathrm{kg}$
• The mass of the second crate, hanging vertically is, $m_2=12 \mathrm{kg}$
• The force on the table is $F=20 \mathrm{N}$
• The friction coefficients of the table is, ${\mu }_s=0.500 \& {\mu }_k=0.325$ 

Step 2: Concept/Significance of work done

Whenever there is the transfer of energy between two bodies, work is said to be done. Thus, work and energy are considered equivalent to each other.

Step 3: (a) Determination of total work done on each one if there is no friction between the tables.

The free-body diagrams of the two blocks are shown below-

                                             

The equation of motion for the block hanging vertically is given by,

$w_2-T=m_{2}a..............................\left(1\right)$ 

The equation of motion for the block sliding horizontally is given by,

$T=m_{1}a..............................\left(2\right)$ 

On comparing both the equations, the acceleration of the blocks is given by,

$$\begin{gathered} a=\frac{w_2}{w_2+m_1} \\ =\frac{w_2}{{\displaystyle \frac{w_1}{g}}+{\displaystyle \frac{w_2}{g}}}..............................\left(3\right) \end{gathered}$$ 

Here, $w_2$ and $w_1$ are the weights of the second and the first block respectively.

For the given values the above equation becomes-

$$\begin{gathered} a=\frac{12 \mathrm{J}}{{\displaystyle \frac{12 \mathrm{J}}{9.8 \mathrm{m}/{\mathrm{s}}^2}}+{\displaystyle \frac{20\mathrm{J}}{9.8 \mathrm{m}/{\mathrm{s}}^2}}} \\ =3.7 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

The work done of the first crate is given by,

$$\begin{gathered} W_1=F·d \\ =m_{1}ad \end{gathered}$$ 

Here, F  is the force applied on the block on the table and d is the distance travelled by the block.

For the given values the above equation gives-

$$\begin{gathered} W_1=\frac{\left(20 \mathrm{J}\right)}{9.8 \mathrm{m}/{\mathrm{s}}^2}\left(3.7 \mathrm{m}/{\mathrm{s}}^2\right)\left(75 \mathrm{cm}\right) \\ =5.66 \mathrm{J} \end{gathered}$$

The work done of the second crate is given by,

$W_2=m_{2}ad$

For the given values the above equation gives-

$$\begin{gathered} W_2=\frac{12 \mathrm{kg}}{9.8 \mathrm{m}/{\mathrm{s}}^2}\times 3.7 \mathrm{m}/{\mathrm{s}}^2\times 0.75 \mathrm{m} \\ =3.4 \mathrm{J} \end{gathered}$$ 

Thus, the work done of the first crate is 5.66 J and the work done of the second crate is 3.4 J when there is no friction.

Step 4: (b) Determination of the work done when and between the table and the 20.0-N block.

In the presence of frictional force $\left(f_k\right)$ , the equation $\left(2\right)$ becomes- 

$$\begin{gathered} T-f_k=m_{1}a \\ T-{\mu }_{k}N=m_{1}a...............................\left(4\right) \end{gathered}$$ 

Here, ${\mu }_k$ is the coefficient of the kinetic frictional force and N is the normal reaction force on the first block, which is equal to the weight of the first block. Thus, from equations , the acceleration on the first block is given as,

$$\begin{gathered} a=\frac{w_2-{\mu }_k{w}_1}{m_1+m_2} \\ =\frac{w_2-{\mu }_k{w}_1}{{\displaystyle \frac{w_1}{g}}+{\displaystyle \frac{w_2}{g}}} \end{gathered}$$ 

Substitute all the values in the above,

$$\begin{gathered} a=\frac{12 \mathrm{N}-0.325\times 20\mathrm{N}}{{\displaystyle \frac{12 \mathrm{N}}{9.8 \mathrm{m}·{\mathrm{s}}^{-2}}}+{\displaystyle \frac{20 \mathrm{N}}{9.8 \mathrm{m}·{\mathrm{s}}^{-2}}}} \\ =1.68 \mathrm{m}·{\mathrm{s}}^{-2} \end{gathered}$$ 

The work done of first crate when friction is present is given by,

$W=F·d$ 

Here, F  is the force applied on the table and d is the distance of table from the top.

Substitute all the values in the above,

$$\begin{gathered} W_1=\frac{\left(20 \mathrm{J}\right)}{9.8 \mathrm{m}/{\mathrm{s}}^2}\left(1.68 \mathrm{m}/{\mathrm{s}}^2\right)\left(75 \mathrm{cm}\right) \\ =2.57 \mathrm{J} \end{gathered}$$ 

The work done of second crate when friction is present is given by,

$W=F·d$ 

Here, F is the force applied on the table and d is the distance of table from the top.

Substitute all the values in the above,

$$\begin{gathered} W_2=\frac{\left(12 \mathrm{J}\right)}{9.8 \mathrm{m}/{\mathrm{s}}^2}\left(1.68 \mathrm{m}/{\mathrm{s}}^2\right)\left(75 \mathrm{cm}\right) \\ =1.54 \mathrm{J} \end{gathered}$$

Thus, the work done by the first crate is 2.57 J and work done by the second crate is 1.54 J when there is friction.