Q : 1. According to work energy theorem for system of particles

(a)

(b)

(c)

(d)

Answer Option : (A)

Explanation :
For a point object 
But system of particles , (just take a bomb) , change of K.E.  Can be done if that bomb explodes (means work done due to internal forces). So for system of particles , .

=> and it is obvious that w_ext , is net external work done and w_int is net internal work done.

Q : 2. If you drop a light ball from height singlequotehsinglequote , and it gets a constant speed after some time

(a) w_grav = Δ K

(b) w_grav + w_resistive = Δ K

(c) w_grav = Δ U

(d) Mechanical energy will remain conserved until it reached the ground

Answer Option : (B)

Explanation :
=> The condition given is of singlequoteterminal speedsinglequote
           
=> As there are two forces on  a ball so it is obvious from work energy theorem that ,
            
=> Mechanical energy will  not remain conserved because there is a resistive frictional force.

=> and w_grav = - Δ U

Q : 3. For the external work done on a earth ball system which one is not possible----->

(a) w_ext = Δ U

(b) w_ext = - w_grav

(c) w_ext = Δ K + Δ U

(d) w_ext = Δ K

Answer Option : (D)

Explanation :

=> options (a), (b), and (c) all are possible .
=> For options (a) and (b) , condition is required .
=> For option (c)  condition is required .
=> option (d) is not possible due to work energy theorem .

Q : 4 . In vertical circular motion

(a) work done by tangential acceleration in a round trip is zero .

(b) total mechanical energy will not remain conserved .

(c) work done  by the tension is half trip is non zero.

(d) none of the above

Answer Option : (A)

Explanation : => In vertical circular motion , there are two components of singlequotemgsinglequote , singlequotemg sin θsinglequote and singlequotemg cos θsinglequote . singlequotemg sin θsinglequote is which cause change in speed (means change in K.E. )and changein K. E. in the round trip is zero so work done  by singlequotemg sin θsinglequote (tangential acceleration) is zero .

=>Tension will always be perpendicular to velocity so work done by tension is zero for any part.

=>Total mechanical energy will remain conserved . Now you can think why ?

Q : 5.


In the above diagram all surfaces are friction - less , pulley , spring and string all are light .
Find how much spring will stretched

(a) 0.3 m

(b)

(c) 0.2 m

(d) 2 m

Answer Option : (C)


Explanation : => Let the extension in the spring is singlequotexsinglequote .
=> Here all the forces singlequotespring forcesinglequote and singlequotegravitational forcesinglequote are conservative so we can apply "conservation of mechanical energy" .
                                        

So,
            

Q : 6.


If we leave the mass m , from the position when spring mass un stretched

(a) Max. displacement of block 

                                               

(b) Max . displacement of block => mg K x₀
                                           

(c) Speed of block at 

(d) All three options (a) , (b) and (c) are wrong.

Answer Option : (A), (C)

Explanation : =>If you leave the block , the block will move down first with increasing speed (because (mg - Kx )>0) and then with decreasing speed (when Kx become greater than mg and (kx - mg)>0) . Finally black will stop.

=> In this condition --->

According to conservation of mechanical energy


=> Speed at equilibrium position will be , you can find it from conservation of mechanical energy.

Q : 7.


The block is at rest w.r. to wedge -->

 Work done by normal reaction in 2 sec.

(a) -20 J

(b) + 20 J

(c) - 48 J

(d) + 48 J

Answer Option : (C)

Explanation :
                

=> From fig.(2) -->
=> From fig.(1) -->
=> If we work from ground frame -->
=>w_f = work done by (f sin θ) , as block is displaced downward and f sin θ is upwards.

=>  x = distance covered by block in 2 sec. (As seen from out side the lift , ground frame).


Q : 8 In the fig shown we have a house of two floors .


(a) The P.E. of ball at first floor must be 'mgh₁'
(b) The P.E. of ball at second floor must be 'mg (h₁ + h₂).

(c) The P.E. of ball at first floor may be zero.

(d) The P.E. of ball at ground level is zero because at ground level h = 0 .

Answer Option : (C)

Explanation :
=> The main concept behind this problem is that you
can consider "any value" of 'P.E.' any where , it is
because the "change in P.E." is important not 'P.E'

=> So, if you consider 'P.E.' at ground level 'zero',the first two options 'may' be correct.

=>So, now you can understand that why only option (c) is correct.

Q : 9.


In the above diagram a block hitsinglequotes a spring and compress it up to a distance singlequotexsinglequote cm . if friction exist from A to B and block leaves the spring at A in negative direction with a speed of 2 m/sec find the compression singlequotexsinglequote in the spring .

(a) 2.5 m

(b) 1.5 m

(c) 2 m

(d) 3 m

Answer Option : (B)

Explanation : =>For the portion 'AB' there will be two forces
will act on block, 'spring' and 'frictional force'.
=>"Spring Force" is a 'Conservative force' and 'Frictional
Force' is a 'non conservative force' .
=> In the question there is a loss of K.E. (mechanical energy) , so that is due to frictional force only .
=> by work energy theorem

 [Because friction will oppose the motion of block while compression of
spring and when spring will try to region it's shape].



Q : 10.


In the above diagram A block is projected upwards with an initial velocity , it just reached the highest position and then came back to lowest point with velocity .

(a)

(b) work done by friction = zero .

(c) work done by mg sin θ is zero

(d) w_grav + w_friction = ΔK

Answer Option : (C) or (D)

Explanation : As shown in the figure friction is present . So Noe you can yourself observe that why options (c) and (d) are correct

Q : 11.If an object is moving with a velocity , so

(a) Its acceleration

(b) Its displacement

(c) Power delivered to it is constant

(d) Options (1) & (2) are wrong .

Answer Option : (B) or (C)

Explanation :



Q : 12.


A train is moving towards right as shown in the figure . The driver of the train sometimes accelerate and sometimes deaccelerate  the train

(a) The work done by kinetic friction on the box is always negative for the above case .

(b) The work done by static friction on the box may be positive and may be negative.

(c) The work done by kinetic friction on the box may be positive for the above case

(d) All options given above correct.

Answer Option : (B) or (C)

Explanation : => When a train accelerate or de accelerate , the box lying on the floor of train may accelerate or decelerate respectively with the help of static friction .

[Ex ---> Box may be at rest w.r.to train.]

=> when the train is accelerating and there is relative motion between train and box, work done by friction on box is positive and when train is decelerating and there is relative motion between train and box , work done by friction on box is negative.

Q : 13. For free foil a height h

(a) K.E. of the body will be half of mgh, at height h/2 from the surface .

(b) Speed of body at height , will be half of the speed when it touches the ground.

(c) T.E. of the earth - body system may be zero at point when ball touch the ground .

(d) All options given above are correct .

Answer Option : (A) or (C)

Explanation : =>Options (A) is correct due to conservation of mechanical energy .

=>Option (C) is correct because you can choose 'any value' of 'P.E.' 'any where'.

=>You can check urself that why rest are wrong.

Q : 14. A simple pendulum is oscillating

(a) The work done by gravitational force may be positive

(b) The work done by gravitational force may be negative

(c) The work done by gravitational force may be zero

(d) The work done by tension will always be zero

Answer Option : (A), (B), (C), or (D)

Explanation : All options are correct

=>The component of weight , 'mg sin θ' will be responsible for increasing and decreasing speed.

=> So work done by it may be positive for short span and may be negative for another short span.

=>For one complete oscillation work done by it will be positive.

=>Tension will always be perpendicular to speed so work done by it is always zero.

Q : 15. If displacement of a particle is zero ,So the work done by a force

(a) will be zero always

(b) will be zero if  is constant

(c) may be non zero.

(d) can not be non zero.

Answer Option : (B) or (C)

Explanation : => Take the example of a constant force .


Net work done by force from .



So options (b) is correct.

Take example of friction force


In the diagram we took an object first from position x₁ ---> x₂ on a frictional floor.

=> From x₁ --> x₂ friction force is against the motion of block, and from x₂ --> x₁, friction force is again against the motion of block .

=> So In the given example even if the displacement of block is zero , the work done b frictional force is not zero , it is negative.

Q : 16.

Work done by friction in 2 sec.

(a) -20 J

(b) + 20 J

(c)  - 36 J

(d) + 36 J

Answer Option : (A)

Explanation :


=> From fig.(2) -->

=> From fig.(1) -->

=> If we observe from ground frame -->

=>w_f = work done by (f sin θ) , as block is displaced downward and f sin θ is upwards.



=>  x = distance covered by block in 2 sec. (As seen from out side the lift , ground frame).



Q : 17.


Answer Option : A:P,Q,R,S,T|B:S,T|C:Q|D:P

Explanation : (A) Linear momentum is conserved in any type of collision. It is conserved even during collision as well . (So all options (P),(Q),(R),(S) and (T) will match).

(B) K.E. remain conserved before elastic collision and after elastic collision (So (S) and (T) will match).

(C) In perfectly Inelastic collision , the loss of K.E. due to collision does not recovered. If due to collision total K.E. lost, in perfectly inelastic collision, no part of the lost K.E recover, So total K.E. may loss in perfectly inelastic collision.

Ex: - If you thro a ball of mud on a wall, it will stick to the wall and will loose total K.E. (So option (Q) is correct)

(D) It is possible in partly inelastic (Inelastic) collision. (So option (P) is correct). 

You might also like following articles