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Question 1
The Potential Energy U in Joule of a particle of mass m=2 kg moving in the x-y plane is given by the formula
$U=6x + 8y+5$
Here x and y are coordinates of the particle in meter
The particle is at (4, 6) at t=0 and v
_{0}=0
i and
j are the unit vectors across x and y axis respectively
(a) Find the force acting on the particle
(b) Find the kinetic energy of the particle at time t
(c) Find the time at which it crosses x axis and y axis and Kinetic energy at that points
(d) Find the coordinates of the particle as a function of time t
(e) Find the Potential Energy of the particle at time t=2 sec
Solution
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
$U=-12-16+5=-23 \J$
Question 2
A body of mass m is displaced from Position A to B by the forces
F_{1} and
F_{2}
Given
Position Vector of Point A=
i+2
j+3
k
Position Vector of Point B=3
i+
j+2
k
F_{1}= i+2
j+3
k
F_{2}= i+
j-2
k
Find the total work done by the resultant forces
Solution
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
Question 3
A block of mass m is placed on a inclined plane of angle α. The Friction Coefficient between the block and surface of the incline plane is μ. The block is placed at height H From the bottom of the incline plane. Find the velocity with which the block reaches the bottom of the incline
Solution
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
$v=\sqrt{2gH(1-\mu c o t{\alpha})}$
Question 4
A block moves horizontally on a rough floor. The Friction Coefficient between the block and surface is μ. The block strikes the light spring of spring constant K with velocity v
(a) Find the maximum compression of the spring
(b) Find the work done by the Friction force
(c) Find the work done by the spring forces
Solution
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$
$=\frac{-\mu^2m^2g^2}{2k}\left[\sqrt{1+\left(\frac{vk}{\mu mg}\right)^2}-1\right]^2$
Question 5
An object of mass M is given an initial velocity v
_{0} on the rough horizontal surface in a positive x –axis direction. The Friction Coefficient between the object and surface is μ.
At t=0, x=0 and i is the unit vector across x axis
(a) Find the instantaneous Power P developed by the Friction forces
(b) Find the velocity of the object at time t
(c) Find the time taken by the object before coming to rest
(d) Find the total work done by the
frictional forces
(e) Find the mean power developed by the frictional forces during the whole motion
(f) if μ varies as per the following
μ=x
Find the value of the x when instantaneous power is maximum
Solution
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
$x=\frac{v_0}{\sqrt{2g}}$
Question 6
A car of mass M accelerates starting from rest along +x axis .The engine is supplying constant Power W .Initially car is at x=0
(a) Find the position of the car as a function of time
(b) Find the velocity of car as a function of time
(c) Find the acceleration of the car as a function of time
Solution
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}$
$=\left(\frac{W}{2Mt}\right)^{1/2}$
Question 7
A uniform chain of mass M and length L is lying on a friction less horizontal table with one fourth of its length hanging over of edge of the table. Find the amount of work done to pull the hanging part of the chain up the table
Solution
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)$
$=\frac{MgL}{32}$
Question 8
A box is dragged across a floor by a rope which makes an angle of 60
^{0} with the horizontal. The tension in the rope is 200 N while the box is dragged is 20 m.
(a) Find the work done
(b) What will be the work done if the rope is horizontal
Solution
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
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