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Work Quiz 2

physics

 Hima applied a force of $8\text{ }N$ in $x-$direction, $3\text{ }N$ in $y-$direction and $2\text{ }N$ in $z-$direction to a body for a time period of $8\,s$.  The displacement because of this force in the $x-$direction be $4\text{ }m$, in $y-$direction be $6\text{ }m$ and in $z-$direction be $8\text{ }m$. What will be the work done by Hima? 

 The energy can be calculated by taking the dot product of force and displacement. The work done will be a scalar quantity even when the force and the displacement are vectors. 

  $\text{The equation of work done,} \\ W=\vec{F}\cdot \vec{s} \\ \text{The force can be given as,} \\ \vec{F}=\hat{i}{{F}_{x}}+\hat{j}{{F}_{y}}+\hat{k}{{F}_{z}} \\ \text{Substituting the values,} \\ \vec{F}=8\hat{i}+3\hat{j}+2\hat{k} \\ \text{The displacement can be shown as,} \\ \vec{s}=x\hat{i}+y\hat{j}+z\hat{k} \\ \text{Substituting the values,} \\ \vec{s}=4\hat{i}+6\hat{j}+8\hat{k} \\ \text{The equation of work,} \\ W=\left( 8\hat{i}+3\hat{j}+2\hat{k} \right)\cdot \left( 4\hat{i}+6\hat{j}+8\hat{k} \right) \\ \text{Substituting the values,} \\ W=8\times 4+3\times 6+2\times 8 \\ \Rightarrow W=32+18+16 \\ \therefore W=66\text{ }J$

 Mohan was playing cricket with Madhu.  Mohan kicked the ball with a force mentioned to be $45\text{ }N$ and it flew away to a distance of $500\text{ }m$ in a direction perpendicular to the direction of the ball initially. What will be the work done by Mohan while batting? 

 The work done can be found by taking the dot product of the force and the displacement.  

  $\text{The equation of work,} \\ W=F\cdot d \\ \Rightarrow W=Fd\cos \theta  \\ \text{Here,} \\ \theta =90{}^\circ  \\ \text{Substituting the values,} \\ W=45\times 500\times \cos 90{}^\circ  \\ \therefore W=0\text{ }J$

 The efficiency of the machine working  for a time period of $1\text{ }hr$ has been mentioned to be as $0.6$. The input power applied by the current source was $30\text{ }W$. What will be the energy obtained from the output? 

 The efficiency will be the ratio of the output power to the input power. Work done can be defined as the product of the power and the time taken. 

 $\text{The efficiency,} \\ \eta =\dfrac{{{P}_{out}}}{{{P}_{in}}} \\ \text{Substituting the values,} \\ \Rightarrow 0.6=\dfrac{{{P}_{out}}}{30} \\ \Rightarrow {{P}_{out}}=18\text{ }W \\ \text{The equation of work done will be,} \\ W=Pt \\ \text{Substituting the values}, \\ W=18\times 3600 \\ \therefore W=64800\text{ }J$

 Somu collected some fluid in a tank and when the tank got filled, he connected a pipe to it with an area of cross section $30\text{ }{{m}^{2}}$. The velocity of fluid flowing through it will be $15\text{ }m{{s}^{-1}}$. The kinetic energy of fluid per sec flowing through the pipe is $60\text{ }J$. Calculate the density of the fluid? 

 The mass can be found by taking the product of the area of the cross section, density of the body, velocity of the body and the time taken. 

 &nbsp;$\text{The equation of kinetic energy,} \\ E=\dfrac{1}{2}m{{v}^{2}} \\ \text{The equation of mass,} \\ m=A\rho vt \\ \text{Substituting the equation of mass in kinetic energy equation,} \\ E=\dfrac{1}{2}A\rho t{{v}^{3}} \\ \Rightarrow \dfrac{E}{t}=\dfrac{1}{2}A\rho {{v}^{3}} \\ \text{Substituting the values,} \\ 60=\dfrac{1}{2}\times \rho \times 30\times {{\left( 15 \right)}^{3}} \\ \text{Rearranging the values}, \\ \Rightarrow \rho =\dfrac{2\times 60}{30\times {{\left( 15 \right)}^{3}}} \\ \therefore \rho =1.18\times {{10}^{-3}}\text{ }kg{{m}^{-3}}$

 Vaishak applied a force of $4{{x}^{2}}\hat{i}+5x\hat{j}+8\hat{k}$ over a body which is moving in a three dimensional space. What will be the energy provided by Vaishak in order to move the body from $\left( 1,1,1 \right)$ to $\left( 5,2,4 \right)$? 

 The volume integral of force and the volume element will give the work done during this application of force. 

 $\text{The force applied,} \\ \vec{F}=4{{x}^{2}}\hat{i}+5x\hat{j}+8\hat{k} \\ \text{As it is a three dimensional space,} \\ \vec{r}=x\hat{i}+y\hat{j}+z\hat{k} \\ \text{The volume element will be,} \\ d\vec{r}=dx\hat{i}+dy\hat{j}+dz\hat{k} \\ \text{The equation of work done,} \\ W=\int{\vec{F}\cdot d\vec{r}} \\ \text{Substituting the values}, \\ W = \int\limits_{\left( 1,1,1 \right)}^{\left( 5,2,4 \right)}{\left( 4{{x}^{2}}\hat{i}+5x\hat{j}+8\hat{k} \right)\cdot \left( dx\hat{i}+dy\hat{j}+dz\hat{k} \right)} \\ \Rightarrow W=\int\limits_{\left( 1,1,1 \right)}^{\left( 5,2,4 \right)}{\left( 4{{x}^{2}}dx+5xdy+8dz \right)} \\ \Rightarrow W=\left[ 4\dfrac{{{x}^{3}}}{3}+5xy+8z \right]_{\left( 1,1,1 \right)}^{\left( 5,2,4 \right)} \\ \Rightarrow W=\left( 166.67+50+32 \right)-\left( \dfrac{4}{3}+5+8 \right) \\ \Rightarrow W=248.67-14.33 \\ \therefore W=234.34\text{ }J$