Q16E - University Physics with Modern Physics

Question

A 1.50-kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s, and at point B it has slowed to 1.25 m/s. (a) How much work was done on the book between A and B? (b) If -0.750 J of work is done on the book from B to C, how fast it is moving at point C? (c) How fast it would be moving at C +0.750 J of work was done on it from B to C?

Step-by-Step Solution

Answer

(a) 6.56 J

 

(b) 0.750 m/s

 

(c) 1.60 m/s

Step 1: Identification of the given data

The given data is listed below as-

• The mass of the book is $m=1.50 \mathrm{kg}$  
• The velocity of sliding book at point A is, $V_1=3.21 m/s$  
• The velocity of sliding book at point B is,  $V_2=1.25 m/s$ 

Step 2: Significance of the work-energy theorem

When forces act on a particle it undergoes displacement, and the particle’s kinetic energy changes by an amount equal to the total work done on the particle by all the forces. Therefore, work done on the particle is given by-

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

The work-energy theorem can be applied to all the bodies that can be treated as particles.

Step 3: Determination of work done on the book between A and B (a)

The work must be done on the book as it changes speed and hence its kinetic energy also changes.

Now, use the work energy theorem,

$W_{net}=K_f-K_i$  

Where, $K=\frac{1}{2}m{V}^2$  

Therefore, work done on the book between A and B is given by-

$$\begin{gathered} W_{net}=K_B-K_A \\ {W}_{net}=\frac{1}{2}m{V_2}^2-\frac{1}{2}m{V_1}^2 \\ {W}_{net}=\frac{1}{2}m\left({V_2}^2-{V_1}^2\right) \end{gathered}$$ 

Here, m is the mass of sliding book, V₁ is velocity of the sliding book at point A and V₂ is the velocity of the sliding book at point B.

 

For, $m=1.50 kg, V_1=3.21m/s, V_2=1.25 m/s$

  

$$\begin{gathered} W_{net}=\frac{1}{2}m\left({V_2}^2-{V_1}^2\right) \\ =\frac{1}{2}\times 1.5 \mathrm{kg}\times \left({\left(1.25\right)}^2 {\mathrm{ms}}^{‐2} -{\left(3.21\right)}^2 {\mathrm{ms}}^{‐2}\right) \\ =6.56 \mathrm{J} \end{gathered}$$

 

Thus, work done on the book between A and B is -6.56 J.

Step 4: Determination of velocity of the book at point C when -0.750 J work is done from point B to C (b)

Now, Initial position of book is B and final position of book is C.

$i=B$ and $f=C$

Therefore, kinetic energy at point C is given by

$$\begin{gathered} K_C=K_B+W_{net} \\ {K}_C=\frac{1}{2}\left(1.50 kg\right){\left(1.25 m/s\right)}^2-0.750 J \\ =+0.422 J \end{gathered}$$  

  

Now, to find velocity at point C, Use  $K_C=\frac{1}{2}m{V_C}^2$

$V_C=\sqrt{\frac{2K_C}{m}}$  

For  $K_C=0.422 J$ and  $m=1.50 \mathrm{kg}$

$$\begin{gathered} V_C=\sqrt{\frac{2\times 0.422 \mathrm{J}}{1.50 \mathrm{kg}}} \\ =0.750 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

Thus, velocity of the book at point C when -0.750 J work is done from point B to C is 0.750 m/s.

Step 5: Determination of velocity of the book at point C when 0.750 J work is done from point B to C (c)

The kinetic energy at point C is given by-

$$\begin{gathered} K_C=K_B+W_{net} \\ {K}_C=\frac{1}{2}\left(1.50 \mathrm{kg}\right){\left(1.25 \mathrm{m}/\mathrm{s}\right)}^2+0.750 \mathrm{J} \\ =+1.922 \mathrm{J} \end{gathered}$$   

  

Now, to find velocity at point C, Use $K_C=\frac{1}{2}m{V_C}^2$ 

$V_C=\sqrt{\frac{2K_C}{m}}$ 

For $K_C=1.922 J$ and  $m=1.50 \mathrm{kg}$

$$\begin{gathered} V_C=\sqrt{\frac{2\times 1.922 \mathrm{J}}{1.50 \mathrm{kg}}} \\ =1.60 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

Thus, velocity of the book at point C when -0.750 J work is done from point B to C is 1.60 m/s.