Now we know that
$\mathbf{F}= \frac{-\partial U}{\partial x}\mathbf{i}+ \frac{-\partial U}{\partial y}\mathbf{j} +\frac{-\partial U}{\partial z}\mathbf{k}$
Here $U=6x + 8y+5$
So,
$\mathbf{F} =-6\mathbf{i}-8\mathbf{j}$
Now $F=ma$
Or $\mathbf{a}=\frac{\mathbf{F}}{m}$
$\mathbf{a}=-3\mathbf{i}-4\mathbf{j}$ ( a constant acceleration)
Now we know that velocity is given by
$\mathbf{v} = \mathbf{u} + \mathbf{a}t$
So here
$v=0-(3\mathbf{i}+4\mathbf{j}) t$
or $v=-(3\mathbf{i}+4\mathbf{j}) t$
Magnitude of velocity
$ =\sqrt{(3t)^2+(4t)^2}$
=5t
Kinetic energy is given by
$ K=\frac{mv^2}{2}$
$ =\frac {2 \times 25t^2}{2}=25t^2$
Now we know that Position vector is given by the equation
$\mathbf{r}=\mathbf{r_0}+\frac{1}{2}\mathbf{a}t^2$
So,
$\mathbf{r}=(4\mathbf{i}+6\mathbf{j})-\frac{1}{2}(3\mathbf{i}+4\mathbf{j})t^2$
Or
$\mathbf{r}=\mathbf{i}(4-\frac{3}{2}t^2)+\mathbf{j}(6-2t^2)$ --(1)
For y axis
$4-\frac{3}{2}t^2=0$
Or$ t=\sqrt{\frac{8}{3}}$
So $KE=25t^2=25 \times 8/3=66.66 \ Joule$
For x axis
$6-2t^2=0$ or $t= \sqrt {3}$
$KE=25t^2=75 \ J$
From equation 1 at time t=2 sec
$\mathbf{r}=-2\mathbf{i}-2\mathbf{j}$
So x= -2 and y=-2
Now $U=6x+ 8y+5$
So U at t=2 sec
$U=-12-16+5=-23 \J$
W=F.S
Where F=Resultant force
S= Displacement
Now here
F=F_{1}+ F_{2}
= (i+2j+3k) + (i+j-2k)
= 2i+3j+k
S=r_{B} – r_{A}
=(3i+j+2k) –( i+2j+3k)
=2i-j-k
So, W=(2i+3j+k).( 2i-j-k)
=4-3-1=0
The below figure shows all the forces acting on the block.
Net force on the block in the inclined direction
$F=mg sin \alpha - \mu mg cos \alpha$
Distance covered across the inclined plane=$\frac {H}{sin \alpha}$
Now from energy equation we know that
$KE_f - KE_i = W$
$\frac{1}{2}mv^2-0=(mgsin{\alpha}-\mu mgcos{\alpha}).\frac{H}{sin{\alpha}}$
Or
$v=\sqrt{2gH(1-\mu c o t{\alpha})}$
Let L be the compression at which block comes to rest
Now,
Net forces acting on the block at any instant during the motion
$ = -kx- \mu mg$
$=-(kx+ \mu mg)$ negative sign indicate it is directed opposite to the motion
Workdone by the net force during the motion=
$=\int_{0}^{L}{-(kx+\mu mg)dx}$
$=-(\frac{1}{2}kL^2+\mu mgL)$
Now from energy equation
$KE_f -KE_i = W$
$0-\frac{1}{2}mv^2=-(\frac{1}{2}kL^2+\mu mgL)$
Solving the quadratic equation and taking positive root
Or
$L=\frac{\frac{-2\mu mg}{k}+\sqrt{\frac{4\mu^2m^2g^2}{k^2}+4v^2}}{2}$
$L=\frac{\mu mg}{k}\left[\sqrt{1+\left(\frac{vk}{\mu mg}\right)^2}-1\right]$
Now Net workdone by the Force of Friction
= $-\mu mgL$
$=\frac{-\mu^2m^2g^2}{k}\left[\sqrt{1+\left(\frac{vk}{\mu mg}\right)^2}-1\right]$
Now net work done by the spring Force
$ =-\frac{1}{2}kL^2$
$=\frac{-\mu^2m^2g^2}{2k}\left[\sqrt{1+\left(\frac{vk}{\mu mg}\right)^2}-1\right]^2$
Force on the object
$\mathbf{F}=- \mu mg \mathbf{i}$
Acceleration
$\mathbf{a}= - \mu g \mathbf{i}$ --(1)
Now velocity
$\mathbf{v}=\mathbf{u}+ \mathbf{a}t$
$\mathbf{v}=v_0 \mathbf{i}- \mu gt \mathbf{i}$
when it will come to rest
$0= v_0\mathbf{i}- \mu gt \mathbf{i} $
Or $t= \frac {v_0}{\mu g}$
Instantaneous Power
$=\mathbf{F}.\mathbf{v}$
$=(- \mu mg \mathbf{i}).( v_0 \mathbf{i}- \mu gt \mathbf{i} )$
$=-\mu mg(v_0-\mu gt)$
Total Work done by the Frictional force=Change in KE
$W= 0 -\frac{1}{2}mv_0^2$
Or
$W=-\frac{1}{2}mv_0^2$
Mean Power developed during whole motion
$=\frac{\int_{0}^{t}Pdt}{t}$
$=\frac{\int_{0}^{v_0/\mu g}{-\mu mg(v_0-\mu gt)dt}}{v_0/\mu g}$
$=\frac{-\mu mg v_0}{2}$
Now when
$\mu =x$
the from equation 1,
$a=-xg$
$\frac{vdv}{dx}=-xg$
Or
vdv=-xgdx
Integrating
$\int_{v_0}^{v}vdv=\int_{0}^{x}{-xgdx}$
Or
$v^2=v_0^2-gx^2$
Or
$v=\sqrt{v_0^2-gx^2}$
Instantaneous Power
$ P=-xmg\sqrt{v_0^2-gx^2}$
For maximum
$\frac{dP}{dx}=0$
Or
$x=\frac{v_0}{\sqrt{2g}}$
Given
W=F.v
$W=M\frac{dv}{dt}v$
Or
Wdt=Mvdv
Integrating
$\int_{0}^{t}Wdt=\int_{0}^{v}Mvdv$
Or
$v=\sqrt{\left(\frac{2Wt}{M}\right)}$
Now
$v=\frac{dx}{dt}$
Or
$vdt=dx$
$\sqrt{\left(\frac{2Wt}{M}\right)}dt=dx$
Integrating
$x=\left(\frac{8W}{9M}\right)^{1/2}t^{3/2}$
Acceleration is defined as
$ a=\frac{dv}{dt}$
$=\left(\frac{W}{2Mt}\right)^{1/2}$
Mass per unit length of the chain=M/L
So mass of the chain hanging over the table=M/4
The center of mass of the hanging part will be located at the middle of the hanging part
=L/8
Now work done in moving it up
$ =\left(\frac{Mg}{4}\right)\left(\frac{L}{8}\right)$
$=\frac{MgL}{32}$
Only the horizontal components of the Tension do the work
$T_x=200 cos 60=100 \ N$
Work done =100*20=2000N
In case rope is horizontal
Work done =200*20=4000N