Q13DQ - University Physics with Modern Physics

Question

A net force acts on an object and accelerates it from rest to speed V₁. In doing so, the force does an amount of work W₁. By what factor must the work done on the object be increased to produce three times the final speed, with the object again starting from rest? 

Step-by-Step Solution

Answer

The work done on the object will increased by a factor of 9 . 

Step 1: About work energy theorem

The total work done is given by:

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

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

Step 2: Determination of the total work done

The total work done using work energy theorem is given by,

  $W_{tot}=K_2‐K_1=∆K$

The Kinetic energy is 0 initially.

$K_{initial}=0$  

Therefore, $W_{tot}=K_{final}$  

Thus, $W_{tot}=\frac{1}{2}m{V^2}_{final}$  

Step 3: Determination of the factor

The work done is given by,

$$\begin{gathered} W_1=\frac{1}{2}m{V^2}_1 \\ {W}_2=\frac{1}{2}m{V^2}_2 \end{gathered}$$ 

Now, the speed is increased 3 times

Therefore, $V_2=2V_1$  

Work is proportional to the square of the speed,

So, $W_2=9W_1$  

Thus, the work should be done nine times to produce three times the final speed and hence the factor is 9 .