Laws of Motion

In the frame of vehicle, vehicle is in equilibrium under the influence of pseudo force F_P

  $F_P=\frac{m{v}^2}{R}$            

N = mg cos 30° + F_P sin 30°

$\Rightarrow N=mgcos30°+\frac{m{v}^2}{R}sin30°.\left(i\right)$              

, and

$f_S=\frac{m{v}^2}{R}cos30°-mgsin30°$              

By doing (1) * cos 30° - (2) * sin 30°, we have

$\Rightarrow N=\frac{800*10}{0.87-0.2*0.5}=\frac{800*10}{0.77}=10.2*10^{3}kgm/s^2$

De Broglie wavelength

$\lambda =\frac{h}{mV}=\frac{h}{\sqrt[]{2mE}}E\to K.E.$              

$E=\frac{3}{2}$  KT for gas

$So\lambda =\frac{h}{\sqrt[]{3mKT}}=\frac{6.6*10^{-34}}{\sqrt[]{3*9*10^{-31}*1.38*10^{-23}*300}}$

$\begin{array}{l}\lambda =6.26*10^{-9}m\\ \lambda =6.26nm\end{array}$                

Suppose acceleration of wedge is a and acceleration of block w. r.t. wedge is a₁ then N cos60° = Ma = 16 a -> N = 32 a

For block w.r.t. wedge

N + 8a sin 30° = 8g cos 30°

N = 8g cos 30° - 8a sin 30°

->32a = 8g cos 30° - 8a sin 30°

->a = $\frac{\sqrt[]{3}}{9}g$  

Now for 8 kg,

$8gsin30°+8acos30°=m{a}_1$  

$a_1=g*\frac{1}{2}+\frac{\sqrt[]{3}}{9}g*\frac{\sqrt[]{3}}{2}$  

$=\frac{2}{3}g$  

For 8 kg block,

8g – 2T = 8a       . (i)

For 2kg block,

T – 2g = 2 * 2a   . (ii)

Solving equations (i) and (ii),

$a=\frac{g}{4}=2.5m/s^2$              

$t=\sqrt[]{\frac{2s}{a}}=\sqrt[]{\frac{2*0.2}{2.5}}=0.4s$              

${\overrightarrow{F}}_x=\left[10*\frac{\sqrt[]{3}}{2}+20*\frac{1}{2}+\frac{20}{\sqrt[]{2}}-\frac{15}{\sqrt[]{2}}-\frac{15\sqrt[]{3}}{2}\right]=9.25\widehat{i}$

${\overrightarrow{F}}_y=\left[15*\frac{1}{2}+20*\frac{\sqrt[]{3}}{2}+10*\frac{1}{2}-\frac{15}{\sqrt[]{2}}-\frac{20}{\sqrt[]{2}}\right]=5\widehat{j}$

Kindly go through the solution 

 $\frac{F_{max}}{1+2}=0.5*10\Rightarrow F_{max}=15N$

  $Tsin\theta =\frac{m{v}^2}{r}$

T cosθ = mg

$tan\theta =\frac{v^2}{rg}$              

$\Rightarrow v=\sqrt[]{rg}$              

F. B. D of Beaker

By NLM2         

$\begin{array}{l}f=ma=m{\omega }^{2}R\\ \Rightarrow m{\omega }^{2}R\le \mu N\\ R\le \mu g/{\omega }^2\end{array}$

  $a=\mu g=0.4*10=4m/s^2$

Slipping stops when V_bag = V_conveyar

-> 0 + at = 2

-> t =    $\frac{2}{a}=\frac{2}{4}=.5sec$

So, x =    $\frac{1}{2}a{t}^2=\frac{1}{2}*4*\frac{1}{4}=0.5m$

  $P=\overrightarrow{F}.\overrightarrow{v}=v\frac{dP}{dt}=v^2\left(\frac{dm}{dt}\right)=0.5*25=12.5W$