alp solutions wpe physics eng

RESONANCE WORK, POWER & ENERGY - 1

TOPIC : WORK, POWER & ENERGYPART - I

1. W = sd.F = )j5it3( . (4t dt i )

= 2

0

2 dtt12=

3t12

20

3 = 32 J

2. For W to be maximum ; dxdW

= 0 ;

i.e. F(x) = 0 x = , x = 0Clearly for d = ,the work done is maximum.Alternate Solution :External force and displacement are in the same direction Work will be positivecontinuously so it will be maximum when displacement is maximum.

3. Work done in changing speed from 0 to V is

W1 = 21

mV2

work done in changing the speed from V to 2V is

W2 = 21

m (2V)2 21

mV2 = 21

3 mV2

21

WW

= 31

4. Applying work energy theorem on block

WF + WS = 0

F 21

k2 = 0 = kF2

or

work done = F = kF2 2

5. In the frame (inertial w.r.t earth) of free end of spring, the initial velocity of block is 3 m/s to left and the springunstretched .

PHYSICS SOLUTIONSADVANCE LEVEL PROBLEMS

TARGET : JEE (IITs)


Applying conservation of energy between initial and maximum extension state.

21

mv2 = 21

kA2 or A = km

v = 3100004

= 6cm.

6. Work done : = Mgh1 + Mgh2 + Mgh3 + 1 Mg1 + 2 Mg2 + 3Mg3 = Mg (h1 + h2 + h3) + Mg (11 + 22 + 33) = Mg (8 + 0.2 + 0.4 + 0.4) = 90 J

7. From conservation of energy

K.E. + P.E. = E or K.E. = E 21

kx2

K.E. at x = kE2

is

E 21

k

kE2

= 0

The speed of particle at x = kE2

is zero.

8. After the top end of chain falls down by 8

, the speed of chain is

v = 8g2

=

2g

.

The mass of chain above table is 87

M.

momentum of chain is

87

M 2g

= 167

M g

9. Let v be the speed of B at lowermost position, the speed of A at lowermost position is 2v.From conservation of energy

21

m (2v)2 + 21

mv2 = mg (2) + mg.

Solving we get v = 5g6

10. Let h be the height of water surface, finally

a2h = a . 2a

. 2a

; h = 4a

C.M. gets lowered by a

8a

4a

= a 8a3

= 8a5

Work done by gravity = mg 8a5


11. It can be observed that power delivered to particle by force F is -P = Fv = K.

The power is constant. Hence work done by force in time t is -W = Pt = Kt

12. The work done by force from time t = 0 to t = t sec. is given by shaded area in graph below.Hence as t increases, this area increases.

Work done by force keeps on increasing.

13. Power P = F

. V

= FV

F = V

dtdm

= V

dtvolume(d

= density

= V

dtvolume(d

= V (AV)= AV2

Power P = AV3or P V3

14. P = dtd (mgh)

Pact =

1000 10 10050

Pact = 2000 W

Pconsumption = 25.0

2000W = 16 kW..

15. When 4 coaches (m each) are attached with engine (2m)according to question P = K 6mgv ..............(1)(constant power), (K being proportionality constant)Since resistive force is proportional to weightNow if 12 coaches are attached

P = K.14mg.v1 ............(2)Since engine power is constantSo by equation (1) and (2)

6Kmgv = 14Kmgv1 v1 = 146

v

= 146

20 = 7106

= 760

= v1 = 8.5 m/secSimilarly for 6 coaches K6mgv = K8mgv2

v2 = 86

20 = 43

20 = 15 m/sec


16. Let F the force with which man pulls the block. Fv = 500 F = 50 N

(F - mg) v = 100solving m = 4 kg

17. At point 'C', the potential energy is minimum, hence it is a point of stable equilibrium.Also, from E to F, the slope is negative i.e.,

F = 0drdU

F is +ve so repulsive

Hence, the force of interaction between the particles is repulsive between points E and F.t

PART - II1. (a) Assume 20 kg and 30 kg block to move together

a = 5050

= 1 m/s2

frictional force on 20 kg block isf = 20 1 = 20 N

The maximum value of frictional force is fmax

= 21

200 = 100 N

Hence no slipping is occurring. The value of frictional force is f = 20 N.Distance travelled in t = 2 seconds.

S = 21

1 4 = 2m.

Work done by frictional force on upper block isW fri = 20 2 = 40 J Ans.

Work done by frictional force on lower block is = 20 2 = 40 J Ans.(b) Yes Ans.(c) Work done by frictional force on the upper block is converted to its kinetic energy. Ans.

2. 12

k3

40

2L

=

12

mv2 +12

k (L0 x)2 when x < L0

v = km

LL x

34

02

02

when x L0 2

04L3K

21

= 2

1mv2 v =

m

k4L3 0

which is also the maximum speed of the block. Thus, vmax

=

34

0L

km

3. (a) For motion to start

4mg5 k

> smg or 5k > 4s


(b)

At the final position of the block extension in spring is maximum and the speed of the block isv = 0. Hence the net work done in taking the block from initial to final positionW = work done by P + work done by spring force F + work done by friction = K = 0

= P x x

0

3 dx.Kx kmgx = 4mg5 k

x 4

Kx4 kmgx = 0

solving we get x = 3/1

KKmg

4. At the moment, elongation is maximum, speed of the block becomes zero.Applying W/E theorem on system

0 21

. mv2 = Wg + Ws

21

. 2.4 = 2.10 . x 2

1 . 100 . x2

Solving x = m513

5. Applying work-energy theorem between A and B.

3m

3m

37

21

mVB2 21

mVA2 = Wgravity + W friction

21

mVB2 21

m (136) = mg(3 + 3 sin 37) mg cos 37 x 3

2

V2B 2

136 = 48 12 VB = 4 m/s


6. Maximum chance of slipping occurs when spring is maximum compressed. At this moment, as force ex-erted by the spring is maximum, acceleration of the system is maximum. Hence maximum friction force isrequired at this moment.By W/E theorem

21 (M + m) V2 = 2

1 Kx

m2 x

m =

KV)mM( 2

Now for upper block am =

mMkxm

force on upper block is provided by the friction force. Therefore mg > mMm.kxm

For limiting value V = g k

mM

using values Vmaximum = 20 cm/s

7.

The speed is maximum when acceleration is leastLet displacement of block is x0 when the speed of block is maximum.At equilibrium, applying Newtons law to the block along the incline

mg sin = mg cos + kx0 ..................(1)Applying work energy theorem to block between initial and final position is

Kf = Ki + mg x0 sin 21

kx02 mg x0 cos ...........(2)Solving (1) and (2) we get,

Vmax

= (sin cos) g km

Ans. Vmax

= (sin cos) g Km

8. (i) Velocity will be maximum when net force = 0.

k.x = N = mg x = mgk

Ws + W f = K

kmg

mgkmgk

21

kmg2k

21 22

= 2

1mv2

On solving

v = kmg

.


(ii) When the particle have velocity equal to zero, then let extension in spring be '

x

'.

12

k 22 mg

k

= mg2 mg

kx

+

12

k

x2

22 2 2m g

k = 2

2 2 2m gk

+ x mg + 12

k x2 x2 mg

kx

= 0

x = 0 (at natural length) or x = 2 mgk

when compression in spring is 2 mg

k i.e. initially

So at natural length, velocity is zero and spring force is also zero. The block will not return or have velocitytowards left.

9. (a) The particle is at equilibrium when F = 0 x (3x 2) = 0

x = 0 and x = 32

m

(b)

The particle is in stable equilibrium at x = 0 metre and unstable equilibrium at x = 32

metre

Since at x = 0 0dxdF

0dxUd2

2

and at x = 32

m 0dxdF

0dx

Ud2

2

(c) The minimum speed imparted to the particle should be such that it just reaches x = 32

from there on

it shall automatically reach x = 0

21

mv2 = 3/2

4

dxF = dx)2x3(x

3/2

4

= 27

1300or v = 27

2600m/s

10. F = x

U

i yU

j = [6 i ] + [8] j = 6 i + 8 j

a

= 3 i + 4 j has same direction as that of

2a

2j4i3

u

| a | = 5 |u | = 5/2Since u

and a

are in same direction, particle will move along a straight line

S = 25

2 + 21

5 22 = 5 + 10 = 15 m.

11. According to W.E. theorem

21

mV2 - 0 = 5

0dx )x410(

V = 10m/sForce at that moment = (10 + 20) = 30 NInstantaneous power = 30 10 = 300W

alp solutions wpe physics eng

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