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This is one of our many quizzes on Power. Click on this to access more such competitive quizzes on Dynamics.

Power Quiz 2

physics

 Arun once pulled a rope with a force mentioned as $5\text{ }N$ which is inclined at an angle given as $\dfrac{\pi }{3}$.  If the power of this force is $3\text{ }W$ which moves at a displacement of $6\,m$, find out the time taken to attain this power.

 Work done can be calculated by taking the dot product of force applied and the displacement happened because of this force. 

  $\text{Work done,} \\ W=F\cdot d \\ \Rightarrow W=Fd\cos \theta  \\ \text{Substituting the values,} \\ W=5\times 6\times \cos \dfrac{\pi }{3} \\ \Rightarrow W=15\text{ }J \\ \text{Work done can be taken as,} \\ W=Pt \\ \Rightarrow t=\dfrac{W}{P} \\ \text{Substituting,} \\ t=\dfrac{15}{3} \\ \therefore t=5\text{ }s$

Somu saw that an electrical wire of resistance $3\text{ M}\Omega $ got heated up when a current of $5\text{ }mA$ passed through it for an interval of $6\text{ }s$. He calculated the power of the current source and obtained an answer as $12.5\text{ }W$. What will be his percentage error in his calculation of power of the current source? 

 The power can be calculated by taking the product of the square of the current and the resistance.  

  $\text{The equation of power,} \\ P={{I}^{2}}R \\ \text{Substituting the values,} \\ P=5\times {{10}^{-3}}\times 5\times {{10}^{-3}}\times 3\times {{10}^{6}} \\ \therefore P=75\text{ }W \\ \text{Percentage error will be,} \\ \%\text{  error}=\dfrac{{{P}_{actual}}-{{P}_{calculated}}}{{{P}_{actual}}}\times 100 \\ \text{Substituting,} \\ \text{% error}=\dfrac{75-12.5}{75}\times 100 \\ \therefore \text{%  error}=83.33\%$

 A body with a value of power mentioned as $5\text{ }W$ took a total time of $100\text{ }s$ to achieve this amount of power. Let us assume that the mass of the body is $1\text{ }kg$. Calculate the momentum of the body 

 The power can be calculated by taking the ratio of the kinetic energy to the time taken. Kinetic energy can be calculated by taking the ratio of the square of the momentum and twice the mass of the body. 

 $\text{The equation of power,} \\ P=\dfrac{\text{Kinetic energy}}{\text{time}} \\ \text{Substituting the values,} \\ 5=\dfrac{E}{100} \\ \Rightarrow E=5\times 100 \\ \therefore E=500\text{ }J \\ \text{The equation of kinetic energy,} \\ E=\dfrac{{{p}^{2}}}{2m} \\ \text{Substituting,} \\ 500=\dfrac{{{p}^{2}}}{2\times 1} \\ \Rightarrow {{p}^{2}}=1000 \\ \Rightarrow p=\sqrt{1000} \\ \therefore p=31.63\text{ kgm}{{\text{s}}^{-1}}$

 Mothika is struggling to find out the power of the body travelling which is represented by the graph below, Calculate the power of the body at $t=4\,s$? <br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/352b7953-56a5-4baf-82f0-dd3286c898038642833479306922432.png"/>

 The area under a force displacement graph will be giving the work done. The power can be calculated by taking the ratio of the work done to the time consumed. 

 $\text{The work done by the body will be the area under the given graph}\text{.} \\ \text{Work done,} \\ W=\text{area of }\Delta \text{ABE+area of rectangle BCDE} \\ \text{Substituting,} \\ W=\dfrac{1}{2}{{b}_{1}}{{h}_{1}}+{{b}_{2}}{{h}_{1}} \\ \Rightarrow W=\left( \dfrac{1}{2}\times 2\times 6 \right)+\left( 3\times 6 \right) \\ \Rightarrow W=6+18 \\ \therefore W=24\text{ }J \\ \text{The equation of power,} \\ P=\dfrac{W}{t} \\ \text{Substituting,} \\ P=\dfrac{24}{4} \\ \therefore P=6\text{ }W$

 Let us assume that a timely changing force mentioned as a two dimensional vector, $\vec{F}=5{{x}^{2}}\hat{i}+3\hat{j}$has been applied on a body. Find out the power of this body if the body is moving from the point $\left( 2,2 \right)$to $\left( 3,4 \right)$ within $4\text{ }s$. 

 The surface integral of the dot product of force and displacement will be given the work done. This work divided by the time taken will provide the value of power. 

 $\text{The force vector,} \\ \vec{F}=5{{x}^{2}}\hat{i}+3\hat{j} \\ \text{The displacement vector in two dimension,} \\ \vec{r}=\left( x\hat{i}+y\hat{j} \right) \\ \text{Therefore,} \\ d\vec{r}=\left( dx\hat{i}+dy\hat{j} \right) \\ \text{Work done,} \\ W=\int{\vec{F}\cdot d\vec{r}} \\ \text{Substituting,} \\ W=\int\limits_{\left( 2,2 \right)}^{\left( 3,4 \right)}{\left( 5{{x}^{2}}\hat{i}+3\hat{j} \right)}\cdot \left( dx\hat{i}+dy\hat{j} \right) \\ \Rightarrow W=\int\limits_{\left( 2,2 \right)}^{\left( 3,4 \right)}{\left( 5{{x}^{2}}dx+3dy \right)} \\ \Rightarrow W=\left[ \dfrac{5{{x}^{3}}}{3}+3y \right]_{\left( 2,2 \right)}^{\left( 3,4 \right)} \\ \Rightarrow W=\left[ 45+12 \right]-\left[ \dfrac{40}{3}+6 \right] \\ \Rightarrow W=57-19.333 \\ \therefore W=37.67\text{ }J \\ \text{Equation of power,} \\ P=\dfrac{W}{t} \\ \Rightarrow P=\dfrac{37.67}{4} \\ \therefore P=9.42\text{ }W$