Work, Power and Energy

CHAPTER-4 WORK, POWER AND ENERGY TRUE/FALSE 1. A simple pendulum with a bob of mass m swings with an angular amplitude of 40°. When its angular displacement is 20°, the tension in the string is greater than mg cos 20°. (1984; 2M)

OBJECTIVE QUESTIONS

positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The toal work done by the force F on the particle is : (1998; 2M) (a) – 2Ka 2 (b) 2 Ka 2 (c) – Ka 2 (d) Ka 2

Only One option is correct : 1. Two masses of 1 g and 4 g are moving with equal kinetic energies. The ratio of the magnitudes of their momenta is : (1980; 2M) (a) 4 : 1

(b)

2:1 (d) 1 : 16

(c) 1 : 2 2.

3.

4.

A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to : (1984; 2M) (a) t1/2 (b) t3/4 (c) t3/2 (d) t2 A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is : (1985; 2M) (a) MgL (b) MgL/3 (c) MgL/9 (d) MgL/18

7.

A spring of force-sonstant k is cut into two pieces such that one piece is double of the other. Then the long piece will have a force-constant of :(1999; 2M) (a) (2/3)k (b) (3/2)k (c) 3 k (d) 6 k

8.

A wind-powered generator converts wind energy into electric energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to: (2000; 2M) (a) v (b) v2 (c) v3 (d) v4

9.

A particle, which is constrained to move along x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F (x) = – kx + ax3 . Here k and a are posiive constant. For x ≥ 0, the functional form of the potential energy U (x) of the particle is : (2002; 2M)

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration a c is varying with time t as a c = k 2 rt2 , where k is a constant. The power delivered to the particle by the force acting on it is : (1994; 1M) (a) 2π mk 2 r2 (b) mk 2 r2 t (c)

( mk 4 r2 t5 ) 3

U (X)

U (X)

(d) zero X

5.

6.

A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is : (1998; 2M)

(a)

(b)

U (X)

(a)

u 2 – 2gL

(b)

2gL

(c)

u 2 – gL

(d)

2 u2 – gL

(

X

U (X)

)

X

X

r

A force F = −K( yiˆ + xjˆ) (where K is a positive constant) acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the 28

(c)

(d)

U (x)

10. An ideal spring with spring-constnat k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is : (2002; 2M) 4 Mg 2 Mg (a) (b) k k (c)

Mg k

(d)

U (x)

x

x

(c)

Mg 2k

11. A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its r acceleration vector a is correctly shown in (2002; 2M)

(d)

14. A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in : (2001; 2M)

→ a

(a)

(b)

→ a

V

V

(a)

(b)

V

→ a (c)

(d)

V

(c)

→ a

12. If W1 , W2 and W3 represent the work done in moving a particle from A to B along different paths, 1, 2 and 3 respectively (as shown) in the gravitational field of a point mass m. Find the correct relation between W1 , W2 and W3 . (2003; 2M)

(d)

15. A bob of mass M is suspended by a massless string of length L. The horizontal velocity v at position A is just sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A, statisfies. (2008; 3M) B

B

2

1

3

θ

A

L

(a) W1 > W2 > W3 (c) W1 < W2 < W3

(b) W1 = W2 = W3 (d) W2 > W1 > W3

13. A particle is placed at the origin and a force F = kx is acting on it (where k is a positive constant). If U (0) = 0, the graph of U (x) versus x will be (where U is the potential energy function : (2004; 2M) U (x)

(c)

U (x)

x

(a)

(a) θ =

x

(b)

π 4

π 3π 0.2 N, block will not move further and it will permanenly stop there. Therefore, total distance covered before it comes to rest permanently is d = AB + BD + x = 2 + 2.14 + 0.1 = 4.24 m (0.2) (0.5) (10) (2.14 + x) =

3. Heat energy required to just melt the bullet. Q = Q1 + Q2 Here, Q1 = ms∆θ = (m × 103 ) (0.03 × 4.2) (327 – 27) = (3.78 × 104 m) J Q2 = mL = (m × 103 ) (6 × 4.2) = (2.52 × 104 m) ∴ Q = (6.3 × 104 )m Only 75% of kinetic energy is utilized to melt the bullet 0.75 ×

B

1 2 mv = Q 2

6.

Let M strikes with v. Then, velocity of m at this instant

1 2 0.75 × × m × v = 6.3 × 104 m ⇒ v = 409.8 m/s 2 4. Normal reaction between blocks A and C will be zero. Therefore, there will be no friction between them. Both A and B are moving with uniform speed. Therefore, net force on them shold be zero.

2

will be v cos θ or

5

v. Further M will fall a distance

of 1 m while m will rise by

(

5–1

)

m. From

conservation of energy principle decrease in potenial energy of M = increase in potenial energy of m + increase in kineic energy of both the blocks.

T=kx 1m 2m

B T=kx

θ

B 1m

f = µ mB g

v θ

m Ag

v cos θ

√5 m

( √5–1) m m

35

1 θ 2

∴

(2) (9.8) (1)

(

= (0.5) (9.8) ⇒ 7.

v

2

1  2v  1 5 – 1 + × 2 × v 2 + × 0.5 ×   2 2  5

)

D

θ

= 3.3 m/s

90° – θ Q mg

(a) At the highest point, velocity of bullet is 50 cos θ. So, by conservation of linear momentum M (50 cos θ) = 4M VA

θ

 50  VA =   cos θ ...(1)  4 At point, B, T = 0 but v ≠ 0

L

∴

VB

L+L sin θ

L 8

u

P

L cos θ

O

We have following conditions,

V=

mv2 L (2) v2 = u 2 – 2gh = u 2 – 2gL (1 + sin θ)

50 cos θ 4

(a) TQ = 0 Therefore, mg sin θ =

2

(4M )V B g 50 ⇒ V B2 = l = Hence, 4 Mg cos 60° = l 2 3 (as l =

(3) QD =

10 m and g = 10 m/s 2 ) 3

3  v2 = V2 – 2g  l  2 

(b)

 50   50  V B2 = V A2 − 100 ⇒   =  cos ?  − 100  3   4  cos θ = 0.86 or θ = 30°

x=

1   cos ? −  = sin 2 θ. cos θ = (1 – cos2 θ) cosθ 8  or, cos ? −

2 Range =  1   u sin2θ  50 × 50 × 3  2   = 2 × 10 × 2 g   2 

= 108.25 m y= H =

u 2 sin 2 ? 50 × 50 ×1 = = 31.25 m 2g 2× 10 × 4

1 = cos ? − cos3 ? 8

3 ∴ cos ? =

1 8

or,cos θ =

1 2

or, θ = 60° From Eq. (1) v2 = gL sin θ = gL sin 60°

Hence, the desired coordinates are (108.25 m, 31.25 m) 8.

...(3)

 v2  Substituting value of  gL  = sin θ from Eq. (1) we get  

2

∴

1 (Range) 2

 v2  1   sin θ cos θ  cos ? −  =   8   gL 

3  V B2 = V A2 − 2 gh ⇒ V B2 = V A2 − 2 g l  2 

or,

...(2)

L v 2 sin2(90° – θ ) v2 sin2θ  = ⇒  L cos ? −  = 8 2g 2g 

By conservation of energy between A and B

or,

...(1)

Let the string slacks at point Q as shown in figure. P to Q path is circular and beyond Q path is parabolic. At point C, velocity of particle becomes horizontal, therefore, QD = half the range of the projectile.

3 gL 2 ∴ Substituting this value of v2 in Eq. (2) u 2 =v2 + 2gL (1 + sin θ) or v2 =

36

=

 3 3 gL + 2 gL 1 +   2 2  

NA

mg

 3 3 3 3 gL + 2 gL = gL  2 +  =  2 2   u=

9.

 3 3 gL  2 +   2  

d  (a) h =  R +  (1–cos θ) 2  Velocity of ball at angle θ θ

2/3

+1

–1

2/3

+1

cos θ

= (3 $j + 3 3 $k ) m/s r r r ∴ V stone = Vstone, cart + Vcart = ( 4iˆ + 3 ˆj + 3 3kˆ) m/s At highest point of its trajectory, the vertical component (z) of its velocity will be zero, whereas the x and ycomponents will remain unchanged. Therefore, velocity of stone at highest point will be, r $ $ V = (4 i + 3 j )m/s or speed at highest point, r V = |V | = ( (4) 2 + (3)2 m/s = 5m/s

mg

d  v2 = 2gh = 2  R +  (1–cos θ) g 2  Let N be the total normal reaction (away from centre) at angle θ. Then,

Substituting value of v2 from Eq. (1) we get mg cos θ – N = 2 mg (1 – cos θ) ∴ N = mg (3 cos θ – 2) (b) The ball will lose contact with the inner sphere when N = 0

cos θ

r 10. (i) Given Vcart = 4iˆ m/s r ∴ V stone, cart = (6 sin30°) $j + (6 cos 30°) $k

v

mv2 mg cos θ – N = d  R +  2 

5mg

2mg

h θ

NB

Now, applying law of conservation of linear momentum, let V0 be the velocity of combined mass after collision. Then, mV = (2m)V0 ∴

V0 =

V 5 m/s = 2.5 m/s = 2 2

(ii) Tension in the string becomes zero at horizontal position. It implies that velocity of combined mass also becomes zero in horizontal position. Applying conservation of energy, we have T=0 V=0

2 or θ = cos–1   3 After this it makes contact with outer sphere and normal reaction starts acting towards the centre. Thus for or, 3 cos θ – 2 = 0

L

2 θ ≤ cos –1   3 NB = 0 and NA = mg (3 cos θ – 2)

V0 = 2.5m/s

2 θ ≥ cos –1   3 NA = 0 and NB = mg (2 – 3 cos θ) The corresponding graphs are as follows

0

and for

37

= V02 – 2gL 2 V02 = (2.5) = 0.32m 2(9.8) 2g

∴

L=

∴

Length of the string is 0.32 m.

11. Let the two blocks move with acceleration a and tension in the string be T.

a=

0. 72 − 0.36 10 × 10 = m/s 2; 0.72 + 0. 36 3

T=

2 × 0.72 × 0. 36 ×10 = 4.8 N; 0. 72 + 0. 36

S=

1 5 × 3. 33 × 12 = m 2 3

WT = 8 J

ASSERATION AND REASON 1.

In statement-I : Decrease in mechanical energ in case I will be

1 2 mv 2 But decrease in mechanical energy in case II will be ∆U 1 =

1 2 mv – mgh 2 ∴ ∆U 2 < ∆U 1 or statement-I is correct. In statement-II : Coefficient of fricition will not change or this statement is wrong. ∴ Option (c) is correct ∆U 2 =

38