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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

Under the acting of a force P, the body can't moves up an incline from initial position $x_1=-150 mm$ to $x_2=80\;m$ where $x=0$ is the unstreched position of the spring along the incline. Find the work done by the spring

\[(a)\;4.03 \;J\quad (b)\;2.25 \;J \quad (c)\; 3.5 \;J \quad (d)\;4.98 \;J\]

1 Answer

From $x=x_1$ to $x=0$ force is along displacement
ie work done $=\large\frac{1}{2}$$kx_1^2$
from $x=0$ to $x=x_2$ force is opposite of displcement so work done is $-\large\frac{1}{2}$$kx_2^2$
$\therefore$ Total work done $=\large\frac{500}{2}$$ \bigg[(0.15)^2-(0.08)^2\bigg]$
$\qquad= 4.03\;J$
Hence a is the correct answer.

 

answered Aug 8, 2013 by meena.p
edited Feb 17, 2014 by meena.p
 

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