Class 11-science H C VERMA Solutions Physics Chapter 8 - Work and Energy

Change in K.E. =  m[v²- u² ]

 =  × 90 × [ 12² - 6² ]

 = 375J

Final speed v = u + at

= 10 + 3 × 5

v = 25m/s

Now,

=    × 2 × 25²

K.E. = 625J

Resisting force

F_R = F. d

F_R = F d cosƟ = 100 ×  4

F_R = 400 J

Work done,

W = mgh

= 5 × 9.8 × (sin30 × 10)

W = 245 J

W = F.d

W = Fd cosƟ

W = 2.5 × 2.5 × cosƟ

W = 6.25J

Now,

v = 28.8 m/s

And F = ma

a =  = 2.5/ 0.015 = 166.66 m/s²

So, v = u+at

t = = 28.8/166.66 = 0.1728 sec

And average power, P_w  = = 6.25/0.1728 = 36.1 W

r = r₂ - r₁

= (3 + 2) - (2 + 3)

r =  -  

Work done W = F × r

W = (5 + 5) × ( - )

W = 0

F = ma

W = F.d

W = 1×40

W =40 J

Work done, W = ₀∫^d

 F dx ( F = a+bx)

W = (a + bd)d

Force = μR and F = mg sin Ɵ 

mg sin Ɵ = μR 

Work done W = μR cos Ɵ 

W = mg sin Ɵ × cos 0 × S

W = 0.25 × 9.8 × 0.60 × 1 × 1

W = 1.5 J

(a) 𝜇_k =  = 

  μ_k=(m + M)g

(b) F_s = μ_kR = (m+M)g × mg

 F_s =  (m + M)

(c) W = F_s d

  W =  (m + M)

(a) R + P sin Ɵ = 2000 ……. 1

P cos Ɵ - 0.2 R = 0……..(2)

Solving 1 and 2 equations we get,

P = 

and work done W = PS cos Ɵ 

W =  

(b) For minimum force,

  [cosƟ + 0.2sinƟ]= 0 (from equation 1)

 Also, W = 40000/5 +tan Ɵ = 40000/5+0.2

 W = 7690 J

Force F = mg sinƟ

F = 100 × sin37⁰ = 60N

(a) Work done W = Fd cosƟ

W = 60 × 2 × cos 0 = 120 J

(b)  In triangle ABC, AB = 2m, Ɵ = 37^o

AC =h = 1.3

work done W = mgh

W = 100 × 1.2

W = 120 J

v² - u² = -2as

a =  

a =  × 25 = 8m/s²

frictional force F = ma = 500 × 8

F =4000 N

v²- u² = 2as

a =  

a =  × 25

a = 8 m/s²

Force, F = ma = 500 × 8

F = 4000N

v = a  

v₁ = 0 and v₂ = a  

v₂² - v²₁ = 2a'd

a^' = =  

Force F = ma^| = (ma²)/2

Work done W = Fd cosƟ

W =  × d

W =  

(a) For a free body,

Force F = mg sinƟ + ma

a=  

a=  

a= 4m/s²

s = ut + at²

s = 0 × 1 +  × 4 × 1²

s = 2m

W = Fs = 20 × 2

W = 40 J

(b) W =Fs

s =  =  

s = 2

W = -mgh = -mg [s sin37⁰]

W = -20 × 1.2

W = -24J

(c) v = u + at = 4 × 10

v = 40 m/s

K.E. = 16 J

(a) S = ut +  at²

S = 0 +  at²

S =5 m

Work done W = Fs cosƟ = 20 × 5 × 1

W = 100 J

(b) h = S sin Ɵ = S sin37⁰ = 3m

work done, W = mgh = 2 × 10 × 3

W =60 J

(c) Frictional force F_f = mg sinƟ

and work done W = F_f s cosƟ

W = mg sinƟ s sinƟ

W = 20 × 0.60 × 5

W = 60 J

We know,

ma = μR 

a =   = μg 

a = 0.1 × 9.8 = 0.98 m/s²

and

v² - u² = 2as

s =   = (0 - 0.4²)/ 2 × 0.98

s = 0.082 m

Work done

W = -μR sinƟ = - 0.1 × 2.5 × 0.082 × 1

W = - 0.02 J

Potential Energy

P.E. = mgh = 1.8 × 10⁵× 9.8 × 50

P.E. = 882 x 10⁵ J/hr

Electrical energy =  P.E. = 441 × 10⁵ J/hr

Power P = (441 × 10⁵)/3600

P = 12.25 × 10³ W

No. of 100W lamps that can be lit = (12.25 × 10³)/100

Number of lamps = 122.5 ≈ 123

Initial P.E. at height h,

P.E_i = mgh = 6 × 9.8 × 2 = 117.6 J

Final P.E. at the floor,

P.E_f = mgh = 0

P.E. = 117.6 - 0

P.E. = 117.6 J

By law of conservation of energy

mgh +  m v_i² =  mv_f

10 × 40 + × 50² =  × v_f²

v =v_f = 57.4 m/s

v ≈ 58 m/s

Power, P = W/t

W = P × t = 460 × 117.56 …..(1)

Also

W = F × d = F × 200 …..(2)

From 1 and 2

F = (460 × 117.56)/200 = 270.3

F = 270 N

(a) v =  = 100/10.54 m/s

K.E. =  mv² =  × 50 × 9.487²

K.E. = 2250 J

(b) Weight =mg = 50 × 9.8 = -490 J

and work done

W = -R d =  × d =  

W = -4900 J

(c) Power P = = 

P = 465 W

 = 30 kg/min = 0.5kg/s

and

Power = × g × h = 0.5 × 9.81 × 10

P = 49 W

Horse power = = 6.6 × 10^-2 hp

Work done W = mgh + mv²

W = 0.2 × 9.8 × 1.5 +( × 0.2 × 3²)

W = 3.84J

W = Fd cosƟ

W = mg cosƟ = 2000 ×10 × 12 × 1

W =24 × 10⁴ J

P == (24 ×10⁴ )/60 = 4000W

Horse power = P/746 = 4000/746 = 5.3 hp

Max. acceleration,

a =  

a = (50/3 - 0)/5 = 3.33m/s²

Force

F = ma = 95 × 3.33

F = 316.6 N

Velocity

v =  = (3.5 × 746)/316.6

v=8.2 m/s

v² - u² = 2as

a = (0.4² - 0)/2 × 2

a = 0.04 m/s²

Force

F = ma - mg

and

Work done

W = Fs cosƟ = m(a-g) s cosƟ

W = 30 (0.04 - 9.8) × 2 × 1

W = -585.5 J ≈ -586 J

Net force is

T - 2mg + 2ma = 0 ….. 1

T - mg - ma = 0….. 2

from equation 1 and 2

a =  =  

a =  m/s²

Now,

s = ut +  at²

s = 0 +  × × 1²

s =  

Decrease in P.E. = mgh

P.E. = (2m - m) × 9.8 × (2/m)

P.E. = 19.6 J

T - 3g + 3a = 0  1

T - 2g - 2a = 0 2

Solving equation 1 and 2

a = m/s²

Now,

S =  (2m - 1)

S =  [2 × 4 -1]

h = S = 6.86 m

Mass

m = m₁ - m₂

m = 3 - 2 = 1kg

Decrease in P.E. = mgh

P.E. = 1 × 9.8 × 6.86

P.E. = 67 J

Work done = change in K.E.

-μR S + mg = [ m₁v₁² m₂v₂²] - 0

-μ × 40 × 2 + 1 × 10 =  (4 × 0.3²+ 1 × 0.6² )

μ = 0.12

Work done by block

= T.E at A - T.E. at B

=  mv²+ mgh -0

= × 0.1 × 5²+ 0.1 × 10 × 0.2

W = 1.45 J

Work done by tube = - Work done by block

W^'= -1.45 J

Work done,

W = K.E_T - P.E_T

W = (0 +  mv²) - mgh

W = [ × 1400 × 15²] - [1400 × 9.8 × 10]

W = 20300J

(a) Work done W = mgh = 0.2 × 10 × 3.2 (to lift the block)

 W = 6.4 J

(b) Work done W = mg sinƟ (to slide the block)

 W = 6.4J

(c) When block falls on ground work done is

 W =  m v² - 0

 6.4 =  × 0.2 × v²

 v = 8m/s

 (d) W =  m v²  - 0

 6.4 =  × 0.2 × v²

 v = 8 m/s

(a) Work done, W = 0 because no work is done by ladder.

(b) Work done W = μRS = fd

= - (200 ×3/10 ) × 10 = -600 J

(c) Work done, W = mg sinƟ × d = 200 × 8/10 × 10

W = 1600 J

By conservation of energy of point A and B

mgH = m v²+ mgh

g × 1 = v²+ g × 0.5

v = √9.8 = 0.3 m/s

After B, body obeys projectile motion, so

S = ut + a t²

-0.5 = u sinƟ t +  (-9.8)t²

t = 0.31 s

and X = 4 cosƟ× t

X =1

Work done, W = mgh = 10 × 1

W = 10 J

Frictional force, F =μR = μmg = 0.2 × 10

F = 2N

and W = Fs

s =  = = 5m

W = (m/l) dx g(x)

Total work done is :

W = ^l/3∫₀  (m/l) dx g(x)

W = mgl/18

Work done for small element is:

dW = μRx 

dw = μ(M/L dx) g(x)

Total work done is :

W = _2L/3∫⁰ μ M/L g (×) dx

W = (2𝜇MgL)/9

Work done = change in P.E.

W = mgh - mgH

W = 1 × 10 × (0.8 - 1)

W = -2 J

k = (mg)/x = 50/0.1 = 500N/m

Total energy T.E. = m v ²+  k x² ……. (1)

and

T.E. (at height h) =  k (h - ×)² + mgh …. (2)

Equating 1 and 2

 m v ²+  k ײ = ½ k (h - x)²+ mgh

× 5 × 2²+  × 500 × 0.1² = × 500 (h - 0.1²) + 5 × 10 × h

h = 0.2 m

h = 20cm

By law of conservation of energy:

 k x² = mgh

 ×100 × 0.1 = mgh = 0.25 × 10 × h

h = 0.2m

h = 20 cm

By work energy principle

(i) for downward motion,

0-0 = (-mg sinƟ × s) - μRs - kx²

80μ + 0.02R = 60 …… (1)

(ii) for upward motion

0 - 0 = (-mg sinƟ × s) - μRs +  kx²

-16 μ + 0.02 k = 12 …… (2)

Solving 1 and 2 we get,

μ = 0.5

and

k = 1000 N/m

Energy at A = Energy at B

 m v ²_a = m v_b²+ k x²

mv² =   + k x²

k = 

From figure

mgx =  k x²

x =  

By law of conservation of energy

m v² =  k₁x²+  k₂x²

So, v = ×√(k₁ + k₂)/m

(a) By law of conservation of energy:

 m v² =  k x²

Max × = v√m/k

(b) Velocity of the block will not be the same when it comes back to the original position, it will be of smaller magnitude and opposite direction.

By law of conservation of energy:

mv² =  k x²

v = 1.58 m/s²

Now, projectile motion

y = (u sinƟ )t -  g t²

Ɵ = 0⁰ , y = -2

t = 0.63 sec

and

x = (u cosƟ)t = 1.58 × 0.63

x = 1m

Minimum velocity at A = v

Minimum velocity at B = 0

By law of conservation of energy

 m v² = mg (2l)

v = 2√gl

from figure

kx cosƟ = mg

40 x X x (0.4/[0.4 + x]) = 0.32 × 10

x = 0.1m

and

S = √(h + x)-h² = √[(0.4+0.1)² - 0.4²]

S = 0.3m

Change in K.E. = Work done

 mv² = -kx²+ mgs

 ×0.32 × v² = - x 40 × 0.1²+0.32 × 10 × 0.3

v =1.5 m/s

cosƟ = cos37^° = BC/AC =4/5

AC = (h + x) =  

and x =AC - h =  - h =  

By work energy principle,

 k x² =  m v²

v = 

v_min = √2gl

Total energy at A = Total energy at B

mgh =  m v² [v = √2gl]

h = l

(a) By law of conservation of energy,

 mv₁² =  m v₂²+ mgl

 × m × (√gl )² =  m v²₂+ mgl 

v₂ =(√8gl)

and

T_B = (mv₂²)/R = [m × (√8gl)²]/l

T_B = 8mg

(b) By law of conservation at A and C

 mv₁² =  m v₃²+ mgh

 × (√logl)² =  × v₃²+ g × 2l

v₃ = (√6gl)

and

 Tension

T_c =[ (mv₃²)/l ] - mg

T_c = 5mg

(c) By law of conservation of energy at A and D

 mv₁² =  m v₄²+ mgh

 (√logl)² =  m v₄²+ gl (1 + cos60⁰)

v₄ = √7gl

and

Tension

T_D = (mv²)/l - mg cos60⁰

by putting v = 7gl we get,

T_D = 6.5mg

from figure,

cosƟ = 

and

AC = AB cosƟ

AC = 0.4

and

CD = AD -AC

CD = 0.1 m

Energy is same at B and D

 m v² = mgh

 v² = 10 × 0.1

v =  m/s

Tension T = (mv²)/r +mg

T= (0.1 ×2)/0.5 + 0.1 × 10

T = 1.4 N

According to figure,

(mv²)/R = mg

v =  

Energy is same at A and P

 kx² =  

x = 

Change in K.E. is given as

 m v² -  mu² =-mgh

v² = 3gl - 2gl (1 +cosƟ) …….. 1

and

(mv²)/l = mg cosƟ

v² =gl cosƟ ……… 2

from equation 1 and 2

3gl - 2gl - 2gl cosƟ = gl cosƟ

θ = cos^-1()

Thus,

Angle = 180⁰ - cos^-1 ()

Angle = cos^-1 (-1/3)

(a) mg cosƟ = (mv²)/l

v = (√gl cosƟ) ….. 1

Change in K.E. is given as :

 mv²  - mu² =mgh

 v² -  (√57)² = g (1.5[ 1 +cosƟ])

v = (√57 - 3g(1 + cosƟ) ……. 2

from equation 1 and 2

1.5 × g cosƟ = 57 - 3g (1+cosƟ)

Ɵ = 53⁰

(b) v = √57 - 3g (1+cosƟ) Ɵ =53

v = 3m/s

(c) After, point B projectile motion will take place

H = OE + DC

H = 1.5 cosƟ + (u²sin²Ɵ)/2×g

H = 1.5 ×  + (9 ×0.8²)/2×10

H = 1.2m

Total energy at A and B are same

K.E_A+ P.E_A = K.E_B+ P.E_B

 we get,

P.E_A = P.E_B

Thus maximum height is same as initial height.

(b)

At C, ½ mv_c² - 0 = mg (L/2)(1 - cos∝)

V_c = (√gL (1-cos∝)) ….. 1

 Again,

V_c = (√gL cos∝) ….. 2

equating 1 and 2

cos∝ = ⅔ …… 3

Now, BF =  + cos∝ 

BF = (5L/6)

(c)

(mv_c²)/L-x = mg

v_c = (√g (L-x)) …. 1

Also,

½ mv_c² - 0 = mg (OC)

v_c  = (√2g (2x-L)) ….. 2

Equating 1 and 2

g (L-x) = 2g (2x - L)

x/L = 0.6

From figure,

(mv²)/R = mg cosƟ

v = (√gRcosƟ) …….. 1

and

Change in K.E. = Work done

 m v² - 0 = mg (R - R cosƟ)

v = (√2gR [1-cosƟ]) ….. 2

from 1 and 2

gR cosƟ = 2gR (1-cosƟ)

Ɵ = cos^-1()

(a) Net force = mg cosƟ = √3/2 mg

(b) N = 0

and

= mg cosƟ 

v = (√gR cosƟ) …..1

and

 mv² = mgR (cos30^° - cosƟ)

v = (√2gR√3/2 - cosƟ]) …. 2

from equation 1 and 2 we get,

Ɵ = cos^-1()

and distance , l =R [Ɵ - π/6]

l = 0.43R

(a) from free body diagram 

N = mg - (mv²)/R

(b) When particle moves with maximum velocity then,

mg = (mv²)/R

v = √gR 

(c ) From figure,

v₁ = (√gR)/2 ……. 1

and

(mv₂²)/R = mg cosƟ

v₂ = (√gR cosƟ) …… 2

Also, change in K.E. = Work done

m v₂² -  m v₁= mgR (1 - cosƟ) …… 3

from equation 2 and 3

gR cosƟ = v₁²+ 2gR (1 - cosƟ)

Ɵ = cos^-1()

(a) F_AB = mg sinƟ

W_B = F_ABS

W_B = mg sinƟ × l ….. 1

and

W_C = mgR(1-cosƟ) ….. 2

Total work done

W = W_B+ W_C

and

change in K.E. = Work done

v₀ = [√2g(R (1-cosƟ) + l sinƟ]

(b) By work energy equation

change in K.E. = Work done

v_c = [√6g (l sinƟ)+R (1-cosƟ) ]

and

Force,

F = (m v_c²)/R

F = 6mg (l/R sinƟ+1-cosƟ)

(c) (mv²)/R = mg cosƟ

v = √gR cosƟ ….. 1

Also,

v = [√2gR (1-cosƟ)]….. 2

equating 1 and 2 we get,

Ɵ = cos^-1 ( )

(a) G.P.E = dm g R cosƟ

= (mgR cosƟ dƟ)/l [dm = m/l R dƟ]

Total G.P.E = ₀∫^l/r(mgR²)/l cosƟ dƟ [Ɵ = l/R]

G.P.E =(mgR²)/l sin(l/R)

(b) K.E. = change in P.E

K.E. = (mgR²)/l [sin (l/R) + sinƟ - sin (Ɵ +l/R)]

(c) From above equation

K.E. = (mgR²)/l [sin (l/R) + sinƟ - sin (Ɵ +l/R)]

as Ɵ = 0

 = (gR/l)[1 - cos(l/R)]

According to figure,

m= ma cosƟ + mg sinƟ

Vdv = a R cosƟ dƟ + gR sin Ɵ dƟ [v = R dƟ/dt]

Integ 

V = [2R (a sin Ɵ + g - g cosƟ)]^½