logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
0 votes

The given graph shows the motion of a particle of mass $10/7\; kg$ the action of an external agency with power. Its initial position is $x=0$ and initial velocity is $1 m/s$ . velocity at $x=10 m$ is (use the graph)

$ a)\; 4 m/s \\ b)\; 2 m/s \\ c)\; 3 \sqrt 2 m/s \\ d)\; 100/3 m/s $

Can you answer this question?
 
 

1 Answer

0 votes
Area under the curve $OABC$
$\quad= \large\frac{1}{2} $$ \times (2+4) \times 10 =30$-----(1)
Also area under the $P-x$ graph
$\quad=\int P dx$
$\quad=\large\int $$\bigg(m \large\frac{dv}{dx}\bigg)$$v dx$
$\quad=\int m \;dv\;v \large\frac{dx}{dt} \qquad [\frac{dx}{dt}$$=v]$
$\quad=\int \limits_1^v m\;v^2\;dv$
$\quad=\large\frac{mv^2}{3} \bigg]_1^v$
$\quad= \large\frac{10}{7 \times 3} $$[v^3-1]$----(2)
From (1) and (2)
$\large\frac{10 }{7 \times 3}$$[v^3-1]=30=>v^3-1=63$
$=>v^3=64$
$=>v=4 m/s$
Hence a is the correct answer. 

 

answered Aug 7, 2013 by meena.p
edited Feb 17, 2014 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...