Solved examples
Question 1. A body fall from height H.if t
_{1} is time taken for covering first half height and
t
_{2} be time taken for second half.Which of these relation is true for t
_{1} and t
_{2}
a. $t_1 > t_2$
b. $t_1 < t_2$
c. $t_1 = t_2$
d. Depends on the mass of the body
Solution.
Let H be the height
then
First Half
$\frac {H}{2}=\frac {1}{2}gt_1^2$ ----(1)
Also $v=gt_1$
Second Half
$\frac {H}{2}=vt_2+\frac {1}{2}gt_2^2$
$\frac {H}{2}=gt_1t_2+\frac {1}{2}gt_2^2$
or
$\frac {1}{2}gt_2^2 =\frac {H}{2}-gt_1t_2$ ---(2)
From 1 and 2
$\frac {1}{2}gt_2^2=\frac {1}{2}gt_1^2 -gt_1t_2$
$t_2^2+2t_1t_2 -t_1^2=0$
or $t_2=\frac {[-2t_1+ \sqrt {(4t_1^2 + 4t_1^2)}]}{2}$
or $t_2=\frac {[-2t_1+2t_1 \sqrt {2}]}{2}$
$t_2=.44t_1$
so, $t_1 > t_2$
Hence a is correct
Question 2. Which of these is true of a conservative force?
a. Work done between two points is independent of the path
b. Work done in a closed loop is zero
c. if the work done by the conservative is positive,its potential energy increases
d. None of the these
Solution. For a conservative force
Work done between two points is independent of the path
Work done in a closed loop is zero
And $-W=U_f-U_i$
So for positive work potential energy decreases
So a and b are correct
Question 3. A simple pendulum consists of a mass attached to a light string l. if the system is oscillating through small angles which of the following is true
a.The frequency is independent of the acceleration due to gravity g
b.The period depends on the amplitude of the oscillation
c.the period is independent of mass m
d. the period is independent of length l
Solution.
Frequency =2π√(l/g)
So it is independent of the mass
Question 4. A body of mass m is dropped from a certain height.it has velocity v
_{1} when it is at a height h
_{1} above the ground.it has velocity v
_{2} when it is at a height h
_{2} above the ground.which of the following is true
a.v
_{1}^{2}-v
_{2}^{2}=2g(h
_{1}-h
_{2})
b.v
_{1}^{2}-v
_{2}^{2}=2g(h
_{2}-h
_{1})
c. v
_{1}-v
_{2}=√2g(√h
_{2}-√h
_{1})
d. v
_{1}-v
_{2}=√2g(√h
_{1}-√h
_{2})
Solution 4
Total Energy at height h
_{1}
=(1/2)mv
_{1}^{2}+2gh
_{1}
Total energy at height h
_{2}
=(1/2)mv
_{2}^{2}+2gh
_{2}
Since we know that total energy remains constant during a free fall
Total Energy at height h
_{1}=Total energy at height h
_{2}
or (1/2)mv
_{1}^{2}+2gh
_{1}=(1/2)mv
_{2}^{2}+2gh
_{2}
or v
_{1}^{2}-v
_{2}^{2}=2g(h
_{2}-h
_{1})
Question 5.A pendulum has a length l.Its bob is pulled aside from its equilibrium position through any angle $\theta$ and then released.The speed of the bob when its passes through it equilibrium position
a. $\sqrt {gl}$
b. $\sqrt {2gl(1-cos \theta)}$
c. $\sqrt {2glcos \theta}$
d.$\sqrt{2gl(1-sin \theta)}$
Solution 5 As shown in figure,the height attained by the bob when the string subtends an angle $\theta$ is
$h=l-lcos \theta$
or $h=l(1-cos \theta)$
So
Potential Energy at this point is given by
$=mgh=mgl(1-cos \theta)$
When the bob passes through equilibrium position,this potential energy is converted into
Kinetic energy
if v be the velocity of the bob then
Kinetic Energy=$\frac {1}{2}mv^2$
Now $\frac {1}{2}mv^2=mgl(1-cos \theta)$
or $v=\sqrt {2gl(1-cos \theta)}$
Question 6.A delivery boy wishes to launch a 2.0 kg package up an inclined plane with sufficient speed to reach the top of the incline.The plane is 3 m long and is inclined at 20 °. Coefficient of friction between the package and the inclined plane is .40. what minimum initial Kinetic Energy must the boy supplied to the package given as sin 20°=.342 cos20°=.940
a. 40 J
b. 42.2 J
c. 42.6 J
d. 45 J
Solution
If the package travels the entire length s of the incline ,the
frictional forces will perform work $-\mu Ns$ where μ is
coefficient of friction
and N is normal reaction.
Let h be the height of the incline plane then the gravitational potential energy of the package will increase by mgh.
Now let assume v speed be given to the package so as to reach the top
Then kinetic energy at the initial point= $\frac {1}{2}mv^2$
Now, applying
Work Energy Theorem
$K_f-K_i=W_{grav} + W_{fric}$
Now since K
_{f}=0
Also Work done by the gravitational force=-(change in gravitational potential energy)=-mgh
Therefore
$-\frac {1}{2}mv^2=-mgh-\mu Ns $
$\frac {1}{2}mv^2=mgh + \mu Ns $
Now s=3
$N=mgcos 20$
$h=s sin 20$
Substituting all the values
$\frac {1}{2}mv^2=42.2J$
Question 7.Choose the correct option
a.if Work done by the conservative force is positive then Potential energy decreases
b. Rate of change of momentum of many particles system is proportional to net external force on the system
c.The work done by the conservative force in closed loop is zero
d. None of the above
Solution 7-
The work done by a conservative force is equal to the negative of the potential energy.When the work done is positive ,the potential energy decreases.
The rate of change of total momentum of a many -particle system is proportional to the net force external to the system ;the internal forces between particles cannot change the momentum of the system.The work done by the conservative system is zero in closed loop
Hence a,b,c are correct
Question 8.The potential energy of a certain particle is given by
$U=20x^2+35z^3$
Find the vector force on it
a. -40x
i-105z
^{2}k
b. 40x
i-105z
^{2}k
c.-10x
i-105z
^{2}k
d. 40x
i+105z
^{2}k
Solution 8.
$U=20x^2+35z^3$
$\mathbf{F}=-\frac{\partial U}{\partial x}\mathbf{i} -\frac{\partial U}{\partial y}\mathbf{j} -\frac{\partial U}{\partial z}\mathbf{k}$
or
F=-40x
i-105z
^{2}k
Matrix Match type
Question 9.
Column I
a. Frictional force
b. Gravitational force
c. Electrical force
d. Viscous force
Column II
P. Work done by the force in closed loop is zero
Q. Work done by the force in closed loop is not zero
Solution
Frictional force is non conservative force
Gravitational force is conservative force
Electrical force is conservative force
Viscous force is non conservative force
And for conservative force, Work done by the force in closed loop is zero
And for non conservative force,Work done by the force in closed loop is not zero
Question 10.Which of the following is non-inertial frame of reference
a. A train which speeding Up
b. A train with constant speed
c. A train which speeding down
d A train at rest
Question 11.what is of these is true for Projectile motion
a.
velocity is perpendicular to
acceleration at the highest point
b. Horizontal components of velocity remains constant through out the path
c. Range of the
Projectile is given by Horizontal component of velocity X Time of flight
d. None of the above
Question 12. A block of mass M is pulled along a horizontal friction surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is
a.Pm/m+M
b.P
c.PM/m+M
d.Pm/M-m
Question 13.A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m/s
^{2}, the reading of the spring balance will be
(A) 24 N
(B) 74 N
(C) 15 N
(D) 49 N
Question 14. A particle moves in a straight line according to
x=t
^{3}-4t
^{2}+3t
Find the acceleration of the particle at
displacement equal to zero
a.(-8,-2,10)
b. (-1,-2,10)
c. (8,2,10)
d. (1,2,10)
Question 14.Which of the following does not have unit as Joule?
a. Work done
b. Kinetic Energy
c. Potential Energy
d. Force
Solution 14
Work done Unit is joule
KE and PE also unit is Joule
Force Unit is Newton
So Solution (d)
Match the column
Question 15
Column A ( Physical quantity)
P)Kinetic Energy
Q)Potential Energy
R)Momentum
S)Mechanical Energy
Column B ( quantities it depends on)
A)Mass
B)Velocity
C)Position of the object
D)Volume
Solution
Kinetic Energy=$\frac {1}{2}mv^2$
Potential Energy=mgh
Momentum=mv
Mechanical Energy= Kinetic Energy + Potential Energy
P -> A,B
Q-> A,C
R-> A,B
S-> A,B.C
Question 16A truck driver pushes the acceleration peddle and increase it speed from v to 2v on the level Road. The mass of the Truck is M. Which of the following is true
a. Potential Energy of the truck does not change
b. Final Kinetic Energy of the truck is $2Mv^2$
c. Work done by the acceleration peddle is $1.5Mv^2$
d. Potential Energy of the truck becomes four times of the initial Potential energy
Solution
Since it is running on level road. PE does not change
Initial Kinetic Energy =$\frac {1}{2}Mv^2$
Final Kinetic Energy=$2Mv^2$
Work done by the acceleration peddle= Final KE - Initial KE=$1.5Mv^2$
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