Search any question & find its solution
Question:
Answered & Verified by Expert
A body is moving up an inclined plane of angle $\theta$ with an initial kinetic energy $E$. The coefficient of friction between the plane and the body is $u$. The work done against friction before the body comes to rest is
Options:
Solution:
2320 Upvotes
Verified Answer
The correct answer is:
$E$
Acceleration of a body up a rough inclined plane
$a=g(\mu \cos \theta+\sin \theta)$
From equation of motion
$\begin{aligned}
v^2 & =u^2-2 a s \\
0 & =u^2-2 a s \\
u^2 & =2 a s
\end{aligned}$
Initial $\mathrm{KE}, E=\frac{1}{2} m u^2$
$\begin{aligned}
E & =\frac{1}{2} m(2 a s) \\
s & =\frac{E}{m a}
\end{aligned}$
$\begin{aligned} & \text { Work done } W=F \cdot s \\ & =(m g \sin \theta+\mu R) \cdot \frac{E}{m a} \\ & =(m g \sin \theta+\mu m g \cos \theta) \cdot \frac{E}{m(\mu \cos \theta+\sin \theta)} \\ & =g(\sin \theta+\mu \cos \theta) \cdot \frac{E}{g(\mu \cos \theta+\sin \theta)}=E\end{aligned}$
$a=g(\mu \cos \theta+\sin \theta)$
From equation of motion
$\begin{aligned}
v^2 & =u^2-2 a s \\
0 & =u^2-2 a s \\
u^2 & =2 a s
\end{aligned}$
Initial $\mathrm{KE}, E=\frac{1}{2} m u^2$
$\begin{aligned}
E & =\frac{1}{2} m(2 a s) \\
s & =\frac{E}{m a}
\end{aligned}$
$\begin{aligned} & \text { Work done } W=F \cdot s \\ & =(m g \sin \theta+\mu R) \cdot \frac{E}{m a} \\ & =(m g \sin \theta+\mu m g \cos \theta) \cdot \frac{E}{m(\mu \cos \theta+\sin \theta)} \\ & =g(\sin \theta+\mu \cos \theta) \cdot \frac{E}{g(\mu \cos \theta+\sin \theta)}=E\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.