Class 11th vectors and work energy and power jee advanced pyq with solutions

WORK, ENERGY AND POWER

Single Type
1. A running man has half the kinetic energy of that of a boy
of half of his mass. The man speeds up by 1 m/s so as to
have same K.E. as that of boy. The original speed of the
man will be
(A) 2 m / s (B) ( 2  1)m / s
(C) 1 m / s (D) 1 m / s
( 2  1) 2

Ans: (C)
Let m = mass of the boy, M = mass of the man, v = velocity
of the boy and V = velocity of the man
Initial kinetic energy of man  21 MV  21  21 m v   21  21  M2  v 
2 2 2

     
 M
m  2 
 
v2 v
 V2   V  ..................................(i)
4 2

When the man speeds up by 1 m/s,
1 1 1M v2
M(V  1)2  m v 2    v 2  (V  1)2 
2 2 2 2  2
v
 V  1 ............................................(ii)
2
1
From (i) and (ii) we get speed of the man V m / s.
2 1




2. A block of mass m is attached to two spring of spring
constant k and k as shown in figure. The block is displaced
1 2




1

, by x towards right and released. The velocity of the block
when it is at its mean position.
k1 k2
m

(k1  k 2 )x 2 3 (k1  k 2 ) x 2
(A) 2m
(B) 4 m
(k1  k 2 ) x 2 (k1  k 2 ) x 2
(C) 4m
(D) m

Ans: (D)



At position A, left spring is extended by x and right spring
is compressed by x:
Applying work energy theorem on block from 'A' to 'B':
WS.F.   K.E.
1
Ui  Uf  mV 2  0
2
1 2 1 2 1
 2 k1x  2 k 2 x   (0  0)  2 mV
2

 
 k  k2  2
 V  1 x
 m 

10
3. A particle A of mass 7
kg is moving in the positive direction
of x. Its initial position is x = 0 & initial velocity is 1
m/s. The velocity at x = 10 is: (use the graph given)




2

, (A) 4 m/s (B) 2 m/s
(C) 3 2 m/s (D) 100
3
m/s

Ans: (A)
Area under P-x graph =
v
dv
v
 mv 3  10
 P dx   mv dt
dx   mv 2 dV  
1
 
 3 1 7  3
(v 3  1)

1
From graph: Area = 2
(2 + 4)×10 = 30

10
 (v 3 –1)=30
73

 v = 4 m/s
ALITER: From graph, P = 0.2 x + 2
or mv dv
dx
v  0.2 x  2

or mv dv =(0.2 x +2)dx
2

v 10
Now integrate both sides,  mv dv   (0.2 x  2) dx
2

1 1




4. In the figure shown, the potential energy U of a particle is
plotted against its position 'x' from origin. Then which of
the following statement is correct. A particle at:




(A) x is in stable equilibrium (B) x is in stable equilibrium
1 2


(C) x is in stable equilibrium (D) none of these
3


Ans: (D)
Condition for equilibrium: F   dU
dx
 0 or
dU
dx
0




3