An object is released from rest from the inner edge of a hemispherical bowl, and it falls under grav...

Let us take : $\phi=\frac{\pi}{2}+\theta$Using work energy conservation, we know that$$ -W_{\text {friction }}=\triangle K E+\triangle P E $$Since both initial and final kinetic energy are the same(0)$$ \begin{gathered} \Delta K E=0 \\ -W_{\text {friction }}=\Delta P E \end{gathered} $$Let the radius of the hemisphere $=R$Then,$$ \Delta P E=m g h=m g R \cos (\phi) $$For any curved surface, the magnitude of frictional force will not be constant so we have to integrate the function of frictionLet us consider a curved surface with $d l$ being small strip where $f=\mu m g \cos \theta$ and $d y=$ horizontal displacment and $d x=$ vertical distanceThen,$$ \begin{gathered} d W=-\mu m g \cos \theta d l=-\mu m g(\cos \theta d l)=-\mu m g d x \\ \int d W=\int-\mu m g d x \\ W=-\mu m g x \end{gathered} $$On substitution,$$ \begin{aligned} \mu m g x & =m g R \cos (\phi) \\ \mu x & =R \cos (\phi) \end{aligned} $$Here,$$ \begin{gathered} x=R(1+\sin (\phi)) \\ \mu R(1+\sin (\phi))=R \cos (\phi) \\ \mu=\frac{\cos (\phi)}{1+\sin (\phi)} \end{gathered} $$On substituting $\phi=\theta-\frac{\pi}{2}=-\left(\frac{\pi}{2}-\theta\right)$$$ \begin{gathered} \cos (\phi)=\cos -\left(\frac{\pi}{2}-\theta\right)=\sin (\theta) \quad \sin (\phi)=\sin -\left(\frac{\pi}{2}-\theta\right)=-\cos (\theta) \\ \mu=\frac{\sin (\theta)}{1-\cos (\theta)} \\ 1-\cos (\theta)=2 \sin ^2 \frac{\theta}{2} \\ \sin \theta=2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right) \\ \mu-\frac{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2 \sin ^2 \frac{\theta}{2}} \\ \mu=\cot \left(\frac{\theta}{2}\right) \end{gathered} $$

A cart of mass $M$ is released from $A$, the highest point of a frictionless track, as shown in the ...

🔹 Step 1: Condition for the cart not to leave the trackAt the top of the loop (point BBB), thecentripetal forceis provided by theweightand thenormal reaction.At the limiting case (just not leaving the track):

Work, Energy And Power - IAT-IISER Physics PYQs

Q: The potential energy of a point particle of mass $m$ undergoing rectilinear motion along the $x$-axi...

The potential energy of a point particle with mass $ m $ moving along the $ x $-axis is described by:$ V(x) = A x + B x^2 $We need to determine the maximum speed attained by the particle if it starts from rest at $ x = \frac{A}{B} $.The particle achieves its maximum speed when its kinetic energy is at its maximum, which occurs when its potential energy is at a minimum. To find where the potential energy is minimized, we calculate the derivative of $ V(x) $ and set it to zero:$ \frac{d(V(x))}{dx} = A + 2Bx = 0 $Solving for $ x $ gives:$ x = -\frac{A}{2B} $The second derivative, $\frac{d^2(V(x))}{dx^2} = 2$, confirms a minimum at this point.Applying the work-energy theorem, we calculate the kinetic energy $ KE_f $ when the potential energy is minimized:Using the initial and final potential energy values:Initial potential energy at $ x = \frac{A}{B} $:$ PE_i = A\left(\frac{A}{B}\right) + B\left(\frac{A}{B}\right)^2 = \frac{A^2}{B} + \frac{A^2}{B} = \frac{2A^2}{B} $Final potential energy at $ x = -\frac{A}{2B} $:$ PE_f = A\left(-\frac{A}{2B}\right) + B\left(\frac{-A}{2B}\right)^2 = -\frac{A^2}{2B} + \frac{A^2}{4B} = -\frac{A^2}{4B} $The change in kinetic energy, using the work-energy theorem:$ KE_f = PE_i - PE_f = \frac{2A^2}{B} - \left(-\frac{A^2}{4B}\right) = \frac{9A^2}{4B} $Expressing kinetic energy in terms of velocity:$ \frac{1}{2} mv^2 = \frac{9A^2}{4B} $Solving for $ v $, the maximum speed:$ v = \sqrt{\frac{9A^2}{2Bm}} = \frac{3A}{\sqrt{2Bm}} $

The potential energy of a point particle with mass

m

moving along the

x

-axis is described by:

V(x)%20%3D%20A%20x%20%2B%20B%20x%5E2

We need to determine the maximum speed attained by the particle if it starts from rest at

x%20%3D%20%5Cfrac%7BA%7D%7BB%7D

.The particle achieves its maximum speed when its kinetic energy is at its maximum, which occurs when its potential energy is at a minimum. To find where the potential energy is minimized, we calculate the derivative of

V(x)

and set it to zero:

%5Cfrac%7Bd(V(x))%7D%7Bdx%7D%20%3D%20A%20%2B%202Bx%20%3D%200

Solving for

x

gives:

x%20%3D%20-%5Cfrac%7BA%7D%7B2B%7D

The second derivative,

%5Cfrac%7Bd%5E2(V(x))%7D%7Bdx%5E2%7D%20%3D%202

, confirms a minimum at this point.Applying the work-energy theorem, we calculate the kinetic energy

KE_f

when the potential energy is minimized:Using the initial and final potential energy values:Initial potential energy at

x%20%3D%20%5Cfrac%7BA%7D%7BB%7D

PE_i%20%3D%20A%5Cleft(%5Cfrac%7BA%7D%7BB%7D%5Cright)%20%2B%20B%5Cleft(%5Cfrac%7BA%7D%7BB%7D%5Cright)%5E2%20%3D%20%5Cfrac%7BA%5E2%7D%7BB%7D%20%2B%20%5Cfrac%7BA%5E2%7D%7BB%7D%20%3D%20%5Cfrac%7B2A%5E2%7D%7BB%7D

Final potential energy at

x%20%3D%20-%5Cfrac%7BA%7D%7B2B%7D

PE_f%20%3D%20A%5Cleft(-%5Cfrac%7BA%7D%7B2B%7D%5Cright)%20%2B%20B%5Cleft(%5Cfrac%7B-A%7D%7B2B%7D%5Cright)%5E2%20%3D%20-%5Cfrac%7BA%5E2%7D%7B2B%7D%20%2B%20%5Cfrac%7BA%5E2%7D%7B4B%7D%20%3D%20-%5Cfrac%7BA%5E2%7D%7B4B%7D

The change in kinetic energy, using the work-energy theorem:

KE_f%20%3D%20PE_i%20-%20PE_f%20%3D%20%5Cfrac%7B2A%5E2%7D%7BB%7D%20-%20%5Cleft(-%5Cfrac%7BA%5E2%7D%7B4B%7D%5Cright)%20%3D%20%5Cfrac%7B9A%5E2%7D%7B4B%7D

Expressing kinetic energy in terms of velocity:

%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%3D%20%5Cfrac%7B9A%5E2%7D%7B4B%7D

Solving for

v

, the maximum speed:

v%20%3D%20%5Csqrt%7B%5Cfrac%7B9A%5E2%7D%7B2Bm%7D%7D%20%3D%20%5Cfrac%7B3A%7D%7B%5Csqrt%7B2Bm%7D%7D

Let us take :

%5Cphi%3D%5Cfrac%7B%5Cpi%7D%7B2%7D%2B%5Ctheta

Using work energy conservation, we know that$

-W_%7B%5Ctext%20%7Bfriction%20%7D%7D%3D%5Ctriangle%20K%20E%2B%5Ctriangle%20P%20E

Since%20both%20initial%20and%20final%20kinetic%20energy%20are%20the%20same(0)

%5Cbegin%7Bgathered%7D%20%5CDelta%20K%20E%3D0%20%5C%5C%20-W_%7B%5Ctext%20%7Bfriction%20%7D%7D%3D%5CDelta%20P%20E%20%5Cend%7Bgathered%7D

Let%20the%20radius%20of%20the%20hemisphere

Then%2C

%5CDelta%20P%20E%3Dm%20g%20h%3Dm%20g%20R%20%5Ccos%20(%5Cphi)

For%20any%20curved%20surface%2C%20the%20magnitude%20of%20frictional%20force%20will%20not%20be%20constant%20so%20we%20have%20to%20integrate%20the%20function%20of%20frictionLet%20us%20consider%20a%20curved%20surface%20with

d l

being%20small%20strip%20where

f=\mu m g \cos \theta

and

d y=

horizontal%20displacment%20and

d x=

vertical%20distanceThen%2C

%5Cbegin%7Bgathered%7D%20d%20W%3D-%5Cmu%20m%20g%20%5Ccos%20%5Ctheta%20d%20l%3D-%5Cmu%20m%20g(%5Ccos%20%5Ctheta%20d%20l)%3D-%5Cmu%20m%20g%20d%20x%20%5C%5C%20%5Cint%20d%20W%3D%5Cint-%5Cmu%20m%20g%20d%20x%20%5C%5C%20W%3D-%5Cmu%20m%20g%20x%20%5Cend%7Bgathered%7D

On%20substitution%2C

%5Cbegin%7Baligned%7D%20%5Cmu%20m%20g%20x%20%26%20%3Dm%20g%20R%20%5Ccos%20(%5Cphi)%20%5C%5C%20%5Cmu%20x%20%26%20%3DR%20%5Ccos%20(%5Cphi)%20%5Cend%7Baligned%7D

Here%2C

%5Cbegin%7Bgathered%7D%20x%3DR(1%2B%5Csin%20(%5Cphi))%20%5C%5C%20%5Cmu%20R(1%2B%5Csin%20(%5Cphi))%3DR%20%5Ccos%20(%5Cphi)%20%5C%5C%20%5Cmu%3D%5Cfrac%7B%5Ccos%20(%5Cphi)%7D%7B1%2B%5Csin%20(%5Cphi)%7D%20%5Cend%7Bgathered%7D

On%20substituting

\phi=\theta-\frac{\pi}{2}=-\left(\frac{\pi}{2}-\theta\right)$$

%5Cbegin%7Bgathered%7D%20%5Ccos%20(%5Cphi)%3D%5Ccos%20-%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D-%5Ctheta%5Cright)%3D%5Csin%20(%5Ctheta)%20%5Cquad%20%5Csin%20(%5Cphi)%3D%5Csin%20-%5Cleft(%5Cfrac%7B%5Cpi%7D%7B2%7D-%5Ctheta%5Cright)%3D-%5Ccos%20(%5Ctheta)%20%5C%5C%20%5Cmu%3D%5Cfrac%7B%5Csin%20(%5Ctheta)%7D%7B1-%5Ccos%20(%5Ctheta)%7D%20%5C%5C%201-%5Ccos%20(%5Ctheta)%3D2%20%5Csin%20%5E2%20%5Cfrac%7B%5Ctheta%7D%7B2%7D%20%5C%5C%20%5Csin%20%5Ctheta%3D2%20%5Csin%20%5Cleft(%5Cfrac%7B%5Ctheta%7D%7B2%7D%5Cright)%20%5Ccos%20%5Cleft(%5Cfrac%7B%5Ctheta%7D%7B2%7D%5Cright)%20%5C%5C%20%5Cmu-%5Cfrac%7B2%20%5Csin%20%5Cleft(%5Cfrac%7B%5Ctheta%7D%7B2%7D%5Cright)%20%5Ccos%20%5Cleft(%5Cfrac%7B%5Ctheta%7D%7B2%7D%5Cright)%7D%7B2%20%5Csin%20%5E2%20%5Cfrac%7B%5Ctheta%7D%7B2%7D%7D%20%5C%5C%20%5Cmu%3D%5Ccot%20%5Cleft(%5Cfrac%7B%5Ctheta%7D%7B2%7D%5Cright)%20%5Cend%7Bgathered%7D

Work, Energy And Power

High-Yield Trend

Questions
4 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Work, Energy And Power

High-Yield Trend

Questions 4 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Questions
4 MCQs