Question 1

Question 10Assertion (A): When the displacement is normal to the direction of force, the work done is zero.Reason (R): Work done depends on the angle between force and displacement.Both A and R are true and R is the correct explanation of ABoth A and R are true but R is not the correct explanation of Aassertion is false but reason is trueassertion is true but reason is false

Accepted Answer

Both A and R are true and R is the correct explanation of A.ExplanationAssertion (A) is true.As,Work done (𝑊) =𝐹⃗\vec𝐹F⋅𝑑⃗\vec𝑑d= Fdcosθ .............(i)WhereF = magnitude of forced = displacementθ = angle between force and displacementIf displacement is normal (perpendicular) to the force, then 𝜃 = 90o90^o90o⇒\Rightarrow⇒ cos 90o90^o90o = 0 ⇒\Rightarrow⇒ W = 0Reason (R) is true.From (i) it is clearly visible that work done depends upon angle between force vector and displacement vector.Hence, R is the correct explanation of A.

Question 2

Question 2A boy of mass 40kg climbs up the stairs and reaches the roof at a height 8m in 5s. Calculate:(i) The force of gravity acting on the boy,(ii)The work done by him against gravity,(iii)The power spent by the boy.Take g = 10 ms-2

Accepted Answer

Given,m = 40 kgh = 8 mt = 5 s(i) Force of gravity acting on the boyF=mgF=40×10=400NF = mg \[0.5em] F = 40 \times 10 = 400N \[0.5em]F=mgF=40×10=400N(ii) Work done against gravityW=F×SW=400×8=3200JW = F × S \[0.5em] W = 400 \times 8 = 3200J \[0.5em]W=F×SW=400×8=3200J(iii)Power spent=work donetime takenW=32005=640W\text{Power spent} = \dfrac{\text{work done}}{\text{time taken}} \[0.5em] W = \dfrac{3200}{5} = 640 WPower spent=time takenwork done​W=53200​=640W

Question 3

Question 3A man spends 7.4 kJ energy in displacing a body by 74 m in the direction in which he applies force, in 2.5 s.Calculate:(i) the force applied and(ii) the power spent (in H.P) by the man.

Accepted Answer

Given,Work done = 7.4 kJS = 74 mt = 2.5 s(i) W = F x SForce=work donedistanceF=7.4×10374⇒F=100N\text{Force} = \dfrac{\text{work done}}{\text{distance}} \[0.5em] F = \dfrac{7.4 \times 10^3}{74} \[0.5em] \Rightarrow F = 100 N \[0.5em]Force=distancework done​F=747.4×103​⇒F=100N(ii)Power spent=work donetime takenP=7.4×1032.5⇒P=2960W\text{Power spent} = \dfrac{\text{work done}}{\text{time taken}} \[0.5em] P = \dfrac{7.4 \times 10^3}{2.5} \[0.5em] \Rightarrow P = 2960 W \[0.5em]Power spent=time takenwork done​P=2.57.4×103​⇒P=2960WWe know,1 H.P=746 W1 W=1746 H.P∴2960 W=2960746 H.P⇒2960 W=3.97 H.P\text{1 H.P} = \text{746 W} \[0.5em] \text{1 W}= \dfrac{1}{746} \text{ H.P} \[0.5em] \therefore \text {2960 W} = \dfrac {2960}{746} \text{ H.P} \[0.5em] \Rightarrow \text {2960 W} = \text{3.97 H.P}1 H.P=746 W1 W=7461​ H.P∴2960 W=7462960​ H.P⇒2960 W=3.97 H.P

Question 4

Question 8A pump is used to lift 600 kg of water from a depth of 75 m in 10 s.Calculate:(a) The work done by the pump,(b) The power at which the pump works, and(c) The power rating of the pump if its efficiency is 40%. (Take g = 10m s-2).

Accepted Answer

Given,Mass of water = 600kgHeight to which the water has to be raised = 75 mTime = 10 s(a) Work done by the pump = mghSubstituting the values we get,W=600×10×75⇒W=4.5×105JW = 600 \times 10 \times 75 \[0.5em] \Rightarrow W = 4.5 \times 10^5 J \[0.5em]W=600×10×75⇒W=4.5×105J(b) Power of the pump=work donetime taken\text{Power of the pump} = \dfrac{\text{work done}}{\text{time taken}}Power of the pump=time takenwork done​(b) ⇒Power=mght\phantom{(b)\space}\Rightarrow\text{Power} = \dfrac{\text{mgh}}{\text{t}}(b) ⇒Power=tmgh​Substituting the values we get,Power=600×10×7510=45000 W⇒Power=45 kW\text{Power} = \dfrac{600 \times 10 \times 75 }{10} = 45000 \text{ W}\[1em] \Rightarrow \text{Power} = 45 \text{ kW} \[1em]Power=10600×10×75​=45000 W⇒Power=45 kW(c) Efficiency=useful powerpower input\text{Efficiency} = \dfrac{\text{useful power}}{\text{power input}}Efficiency=power inputuseful power​Given,Efficiency = 40 %0.4=45input powerinput power=45×1030.4=112.5×103 W=112.5 kW0.4 = \dfrac{45}{\text{input power}} \[0.5em] \text{input power} = \dfrac{45 \times 10^3}{0.4} \[0.5em] = 112.5 \times 10^3 \text{ W} \[0.5em] = 112.5 \text{ kW}0.4=input power45​input power=0.445×103​=112.5×103 W=112.5 kW∴ Power rating of pump = 112.5 kW.

Question 5

Question 15A man raises a box of mass 50kg to a height of 2m in 20s, while another man raises the same box to the same height in 50s.(a) Compare: (i) the work done, (ii) the power developed by them.(b) Calculate the (i) the work done, (ii) the power developed by each man.Take g = 10N kg-1.

Accepted Answer

(a)(i) Work done = force x displacementHence, work done is same for both men as both carry 50 kg weight to a height of 2m.Therefore,Work done by first man : Work done by second man = 1 : 1(ii)Let the work done by both men be W.Power=work donetimePower developed by Man APower developed by Man B=work done by Man Atime taken by Man Awork done by Man Btime taken by Man B=W20W50=52\text{Power} = \dfrac{\text{work done}}{\text{time}} \[1em] \dfrac{\text{Power developed by Man A}}{\text{Power developed by Man B}} \[1em] = \dfrac{\dfrac{\text{work done by Man A}}{\text{time taken by Man A}}}{\dfrac{\text{work done by Man B}}{\text{time taken by Man B}}} \[1em] = \dfrac{\dfrac{W}{20}}{\dfrac{W}{50}} \[1em] = \dfrac{5}{2}Power=timework done​Power developed by Man BPower developed by Man A​=time taken by Man Bwork done by Man B​time taken by Man Awork done by Man A​​=50W​20W​​=25​Power developed by Man A : Power developed by Man B = 5:2(b)(i) As we know,W = F x S = mghW=50×10×2=1000J⇒W=1000JW = 50 \times 10 \times 2 = 1000 J \[0.5em] \Rightarrow W = 1000 J \[0.5em]W=50×10×2=1000J⇒W=1000J(ii) Power developed by Man AP=100020⇒P=50WP = \dfrac{1000}{20} \[0.5em] \Rightarrow P = 50 W\[0.5em]P=201000​⇒P=50WPower developed by Man BP=100050⇒P=20WP = \dfrac{1000}{50} \[0.5em] \Rightarrow P = 20 W\[0.5em]P=501000​⇒P=20W

Question 6

Question 16A boy takes 3 minutes to lift a 20 litre water bucket from a 20 m deep well, while his father does it in 2 minutes.(a) Compare:(i) the work, and (ii) power developed by them.(b) How much work each does?Take density of water = 103kg m -3 and g = 9.8 N kg-1

Accepted Answer

(a)(i) Work done = force x displacementWork done is same for boy and father as both lift 20 litre water from 20 m deep well.Work done by boy : Work done by father = 1:1(ii) Comparing Power developed:Power=work donetimePower of boy:Power of father =W3W2⇒Power developed by boy:Power developed by father=23\text{Power} = \dfrac{\text{work done}}{\text{time}} \[0.5em] {\text {Power of boy}} : {\text {Power of father }} = \dfrac{\dfrac{W}{3}}{\dfrac{W}{2}} \[0.5em] \Rightarrow {\text {Power developed by boy}} : {\text {Power developed by father}} = \dfrac{2}{3} \[0.5em]Power=timework done​Power of boy:Power of father =2W​3W​​⇒Power developed by boy:Power developed by father=32​Power developed by boy : Power developed by father = 2:3(b) Work done = mghGiven,density of water = 103kg m -3To convert, density of water to mass per litreWe know,1 litre = 1000 cm3and1 m3 = 106 cm3So,103 kgm−3=1000 kg1 m3=1000 kg106 cm3=1 kg1000 cm3=1 kg1 litre10^3\text{ kgm}^{-3} = \dfrac{1000\text{ kg}}{1\text{ m}^3} \[1em] = \dfrac{1000\text{ kg}}{10^6 \text{ cm}^3} \[1em] = \dfrac{1\text{ kg}}{1000\text{ cm}^3} \[1em] = \dfrac{1\text{ kg}}{1 \text{ litre}}103 kgm−3=1 m31000 kg​=106 cm31000 kg​=1000 cm31 kg​=1 litre1 kg​Hence,20L of water = 20 Kg by massheight = 20mNow, substituting the values in the formula for work done we getW = 20 x 9.8 x 20= 3.92 kJ

Question 7

Question 14Assertion (A): If momentum of a body increases by 50%, its kinetic energy will increase by 125%.Reason (R): Kinetic energy is proportional to the square of velocity.Both A and R are true and R is the correct explanation of ABoth A and R are false and R is not the correct explanation of Aassertion is false but reason is trueassertion is true but reason is false

Accepted Answer

Both A and R are true and R is the correct explanation of A.ExplanationAssertion (A) is true. When the momentum of a body increases by 50%, its kinetic energy will increase by 125%.Let initial momentum be p and kinetic energy be KIf p increases by 50%, the new momentum p' = p + p2\dfrac{p}{2}2p​ = 3p2\dfrac{3p}{2}23p​Relation between kinetic energy and momentum is given by; K = p22m\dfrac{p^2}{2m}2mp2​So, K′K\dfrac{K'}{K}KK′​ = p′22mp22m\dfrac{\dfrac{p'^2}{2m}}{\dfrac{p^2}{2m}}2mp2​2mp′2​​ = p′2p2\dfrac{p'^2}{p^2}p2p′2​ = (3p2)2p21\dfrac{\Big(\dfrac{3p}{2}\Big)^2}{\dfrac{p^2}{1}}1p2​(23p​)2​ = 94\dfrac{9}{4}49​So,Percentage increase in kinetic energy = K′−KK\dfrac{K'-K}{K}KK′−K​ x 100= 9K4−KK\dfrac{ \dfrac{9K}{4} -K}{K}K49K​−K​ x 100= 9K−4K4K\dfrac{ \dfrac{9K-4K}{4}}{K}K49K−4K​​ x 100= 5K4K\dfrac{ \dfrac{5K}{4}}{K}K45K​​ x 100= 54\dfrac{5}{4}45​ x 100 = 125%Hence, kinetic energy will increase by 125%.Reason (R) is true. Kinetic energy is proportional to the square of velocity, according to the formula: KE = 12\dfrac{1}{2}21​mv2.

Question 8

Question 9A block A, whose weight is 100N, is pulled up a slope of length 5m by means of a constant force F (=100N) as illustrated.(a) What is the work done by the force F in moving the block A, 5m along the slope?(b) What is the increase in potential energy of the block A?(c) Account for the difference in the work done by the force and the increase in potential energy of the block.

Accepted Answer

Given,F = 100 Nh = 3mdisplacement = 5mWe know,(a) Work done = Force × displacement in the direction of the forceW=100×5⇒W=500JW = 100 \times 5 \[0.5em] \Rightarrow W = 500 J \[0.5em]W=100×5⇒W=500J(b) The potential energy gained by the block U = mghP.E.=100×3⇒ P.E. gained by the block=300J\text{P.E.} = 100 \times 3 \[0.5em] \Rightarrow \text { P.E. gained by the block}= 300 J \[0.5em]P.E.=100×3⇒ P.E. gained by the block=300J(c) The difference (200 J) energy is used in doing work against the force of friction between the block and the slope and it will appear as heat energy.

Question 9

Question 11If the speed of a car is halved, how does its kinetic energy change?

Accepted Answer

Let speed of car be v.Kinetic Energy at speed v = 12\dfrac{1}{2}21​ x m x v2 = mv22\dfrac{	ext{mv}^2}{2}2mv2​Kinetic Energy at speed v2\dfrac{	ext{v}}{2}2v​ = 12\dfrac{1}{2}21​ x m x (v2)2(\dfrac{	ext{v}}{2})^2(2v​)2 = mv28\dfrac{	ext{mv}^2}{8}8mv2​ = 14(mv22)\dfrac{1}{4}\Big(\dfrac{	ext{mv}^2}{2}\Big)41​(2mv2​)Hence, when the speed of the car is halved the kinetic energy becomes one-fourth.

Question 10

Question 12Two bodies of equal masses are moving with uniform velocities v and 2v. Find the ratio of their kinetic energies.

Accepted Answer

Given,Velocity of first body v1 = vVelocity of second body, v2 = 2vAs the mass of the two bodies are same and as kinetic energy is directly proportional to the square of the velocity (K α v2)Hence, ratio of their kinetic energies isK1K2=(v1)2(v2)2K1K2=v2(2v)2K1K2=v24v2K1K2=14\dfrac{K_1}{K_2} = \dfrac{(v_1)^2}{(v_2)^2} \[0.5em] \dfrac{K_1}{K_2} = \dfrac{v^2}{(2v)^2} \[0.5em] \dfrac{K_1}{K_2} = \dfrac{v^2}{4v^2} \[0.5em] \dfrac{K_1}{K_2} = \dfrac{1}{4} \[0.5em]K2​K1​​=(v2​)2(v1​)2​K2​K1​​=(2v)2v2​K2​K1​​=4v2v2​K2​K1​​=41​

Question 11

Question 13Two bodies have masses in the ratio 5 : 1 and kinetic energies in the ration 125 : 9. Calculate the ratio of their velocities.

Accepted Answer

We know that,K.E.=12×m×v2\text{K.E.} = \dfrac{1}{2} \times m \times v^2 \[0.5em]K.E.=21​×m×v2Let the two bodies be A and B.Let mass of body A be mA and body B be mB. Let velocity of body A be vA and body B be vB.Given,Ratio of masses:mAmB=51\dfrac {m_A}{m_B} = \dfrac {5}{1} \[0.5em]mB​mA​​=15​Ratio of K.E.:KEAKEB=1259KEAKEB=12×mA×vA212×mB×vB2⇒1259=12×mA×vA212×mB×vB2⇒1259=mAmB×(vAvB)2⇒1259=51×(vAvB)2⇒(vAvB)2=1259×15⇒(vAvB)2=259⇒vAvB=259⇒vAvB=53\dfrac {KE_A}{KE_B} = \dfrac {125}{9} \[1em] \dfrac {KE_A}{KE_B} = \dfrac{\dfrac{1}{2} \times m_A \times v_A^2}{\dfrac{1}{2} \times m_B \times v_B^2} \[0.5em] \Rightarrow \dfrac {125}{9} = \dfrac{\dfrac{1}{2} \times m_A \times v_A^2}{\dfrac{1}{2} \times m_B \times v_B^2} \[0.5em] \Rightarrow \dfrac {125}{9} = \dfrac{m_A}{m_B} \times \Big(\dfrac{v_A}{v_B}\Big)^2 \[0.5em] \Rightarrow \dfrac {125}{9} = \dfrac{5}{1} \times \Big(\dfrac{v_A}{v_B}\Big)^2 \[0.5em] \Rightarrow \Big(\dfrac{v_A}{v_B}\Big)^2 = \dfrac{125}{9} \times \dfrac{1}{5} \[0.5em] \Rightarrow \Big(\dfrac{v_A}{v_B}\Big)^2 = \dfrac{25}{9} \[0.5em] \Rightarrow \dfrac{v_A}{v_B} = \sqrt{\dfrac{25}{9}} \[0.5em] \Rightarrow \dfrac{v_A}{v_B} = \dfrac{5}{3} \[0.5em]KEB​KEA​​=9125​KEB​KEA​​=21​×mB​×vB2​21​×mA​×vA2​​⇒9125​=21​×mB​×vB2​21​×mA​×vA2​​⇒9125​=mB​mA​​×(vB​vA​​)2⇒9125​=15​×(vB​vA​​)2⇒(vB​vA​​)2=9125​×51​⇒(vB​vA​​)2=925​⇒vB​vA​​=925​​⇒vB​vA​​=35​∴ Ratio of their velocities = 5:3

Question 12

Question 17A body of mass 10 kg is moving with a velocity 20 m s-1.If the mass of the body is doubled and its velocity is halved,find:(i) the initial kinetic energy, and(ii) the final kinetic energy.

Accepted Answer

Given,Initial Mass = 10 kg andFinal mass = 2 × 10 kg = 20kgInitial Velocity = 20 m / sChanged velocity v2 = 20 / 2 = 10 m / s(i) Initial kinetic energy = 1 / 2 × mass × (velocity)2Initial Kinetic Energy=12×10×(20)2⇒Initial kinetic energy=2000J\text {Initial Kinetic Energy} = \dfrac{1}{2} \times 10 \times (20)^2 \[0.5em] \Rightarrow \text{Initial kinetic energy} = 2000J \[0.5em]Initial Kinetic Energy=21​×10×(20)2⇒Initial kinetic energy=2000J(ii) Final kinetic energyFinal Kinetic Energy=12×20×(10)2⇒Final kinetic energy=1000J\text {Final Kinetic Energy} = \dfrac{1}{2} \times 20 \times (10)^2 \[0.5em] \Rightarrow \text{Final kinetic energy} = 1000J \[0.5em]Final Kinetic Energy=21​×20×(10)2⇒Final kinetic energy=1000J

Question 13

Question 18A truck weighing 1000 kg changes its speed from 36 km h-1 to 72 km h-1 in 2 minutes.Calculate :(i) the work done by the engine and(ii) its power(g =10 m s-2)

Accepted Answer

Given,Mass = 1000 kgInitial velocity (u) =36km/h=36×1000m3600s⇒Initial velocity (u) =10ms−1\text {Initial velocity (u) } = 36 km / h \[0.5em] = \dfrac {36 \times 1000m}{3600s} \[0.5em] \Rightarrow \text {Initial velocity (u) } = 10ms^{-1} \[0.5em]Initial velocity (u) =36km/h=3600s36×1000m​⇒Initial velocity (u) =10ms−1Final velocity (v) =72km/h=72×1000m3600s⇒Final velocity (v) =20ms−1\text {Final velocity (v) } = 72 km / h \[0.5em] = \dfrac {72 \times 1000m}{3600s} \[0.5em] \Rightarrow \text {Final velocity (v) } = 20ms^{-1} \[0.5em]Final velocity (v) =72km/h=3600s72×1000m​⇒Final velocity (v) =20ms−1From, equation of motion we gets=(v2−u2)/2as = ( v^2 - u^2) / 2as=(v2−u2)/2a(i) Work done = force x displacementW=F.s=ma.s=1000×(202−102)2W=1000×(400−100)2W=500×300⇒W=150000J=1.5×105JW = F.s = ma.s = \dfrac{1000 \times (20^2 -10^2)}{2} \[0.5em] W = \dfrac{1000 \times (400 -100)}{2} \[0.5em] W = 500 \times 300 \[0.5em] \Rightarrow W = 150000J = 1.5 × 10^5J \[0.5em]W=F.s=ma.s=21000×(202−102)​W=21000×(400−100)​W=500×300⇒W=150000J=1.5×105J(ii) Power = work done / time takenPower=1.5×105120⇒Power=1.25×103WPower = \dfrac{1.5 \times 10^5}{120} \[0.5em] \Rightarrow Power = 1.25 × 10^3 WPower=1201.5×105​⇒Power=1.25×103W

Question 14

Question 19A body of mass 60 kg has the momentum of 3000 kgms-1.Calculate:(i) the kinetic energy and(ii) the speed of the body.

Accepted Answer

Given,Mass = 60 kgMomentum = 3000 kgm / s(i)Kinetic Energy=momentum22×mass⇒KE=300022×60⇒KE=3000×3000120⇒KE=75000 J=7.5×104 J\text{Kinetic Energy} = \dfrac{\text{momentum}^2}{2 \times \text{mass}} \[0.5em] \Rightarrow \text{KE} = \dfrac{3000^2}{2 \times 60} \[0.5em] \Rightarrow \text{KE} = \dfrac{3000 \times 3000}{120} \[0.5em] \Rightarrow \text{KE} = 75000\text{ J} = 7.5 × 10^4\text{ J}Kinetic Energy=2×massmomentum2​⇒KE=2×6030002​⇒KE=1203000×3000​⇒KE=75000 J=7.5×104 J\(ii)Momentum=Mass×Velocity⇒Velocity=MomentumMass⇒Velocity=300060⇒Velocity=50 ms−1\text{Momentum} = \text{Mass} \times \text{Velocity} \[0.5em] \Rightarrow \text{Velocity} = \dfrac{\text{Momentum}}{\text{Mass}} \[0.5em] \Rightarrow \text{Velocity} = \dfrac{3000}{60} \[0.5em] \Rightarrow \text{Velocity} = 50\text{ ms}^{-1}Momentum=Mass×Velocity⇒Velocity=MassMomentum​⇒Velocity=603000​⇒Velocity=50 ms−1

Question 15

Question 20How much work is needed to be done on a ball of mass 50g to give it a momentum of 5 kg m s-1?

Accepted Answer

Given,mass = 50gm = 0.05kgmomentum = 500 gcm / s = 0.005 kgm / sKE = momentum2 / 2 x massWork done=KE=momentum22×massWork done=KE=0.00522×0.05⇒KE=250J\text{Work done} = KE = \dfrac {momentum^2}{2 \times mass} \[0.5em] \text{Work done} = KE = \dfrac{0.005^2} {2 \times 0.05} \[0.5em] \Rightarrow KE =250 J \[0.5em]Work done=KE=2×massmomentum2​Work done=KE=2×0.050.0052​⇒KE=250J

Question 16

Question 21How much energy is gained by a box of mass 20 kg when a man(a) carrying the box waits for 5 minutes for a bus?(b) runs carrying the box with a speed of 3 ms-1 to catch the bus?(c) raises the box by 0.5 m in order to place it inside the bus?(g=10 ms-2)

Accepted Answer

Given,Mass = 20 kg(a) As there is no displacement of the man, so work done is Zero.(b) We know,Work done = Kinetic energy of man = 1 / 2 × mass × (velocity)2Work done=Kinetic energy of man=12×mass×velocity2=12×20×32=12×20×9=90J\text{Work done} = \text{Kinetic energy of man} \[0.5em] = \dfrac{1}{2} \times \text{mass} \times \text{velocity}^2 \[0.5em] = \dfrac{1}{2} \times 20 \times 3^2 \[0.5em] = \dfrac{1}{2} \times 20 \times 9 \[0.5em] = 90 J \[0.5em]Work done=Kinetic energy of man=21​×mass×velocity2=21​×20×32=21​×20×9=90J∴ Energy Gained = 90 J(c) We know,Potential energy = mghPE=20×10×0.5⇒U=100J\text{PE} = 20 \times 10 \times 0.5 \[0.5em] \Rightarrow U = 100 J \[0.5em]PE=20×10×0.5⇒U=100J

Question 17

Question 22A bullet of mass 50g is moving with a velocity of 500ms-1. It penetrates 10 cm into a still target and comes to rest.Calculate:(a) the kinetic energy possessed by the bullet, and(b) the average retarding force offered by the target.

Accepted Answer

Given,Mass = 50 g = 0.05 kgVelocity = 500 m / sDistance (S) = 10 cm = 0.1 m(a) The kinetic energy possessed by the bullet = 12\dfrac{1}{2}21​ × mass × (velocity)2Substituting the values in equation we get,KE=12×0.05×5002KE=12×0.05×250000⇒KE=6250JKE = \dfrac {1}{2} \times 0.05 \times 500^2 \[0.5em] KE = \dfrac {1}{2} \times 0.05 \times 250000 \[0.5em] \Rightarrow KE = 6250J \[0.5em]KE=21​×0.05×5002KE=21​×0.05×250000⇒KE=6250J(b)Retarding Force=Work done by bulletdistance=62500.1=62500N\text{Retarding Force} = \dfrac{\text{Work done by bullet}}{\text{distance}} \[0.5em] = \dfrac{6250}{0.1} \[0.5em] = 62500N \[0.5em]Retarding Force=distanceWork done by bullet​=0.16250​=62500N∴ Average retarding force offered by target = 62500N

Question 18

Question 23A spring is kept compressed by a small trolley of mass 0.5 kg lying on a smooth horizontal surface as shown. When the trolley is released, it is found to move at a speed of v = 2 ms-1. What potential energy did the spring possess when compressed?

Accepted Answer

Given,Mass = 0.5 kgVelocity = 2 m / sPotential Energy possessed by the spring is equal to the kinetic energy of moving trolley.Kinetic energy of trolley=12×mass×velocity2=12×0.5×22=12×0.5×22=1 J= \dfrac{1}{2} \times \text{mass} \times \text{velocity}^2 \[0.5em] = \dfrac{1}{2} \times 0.5 \times 2^2 \[0.5em] = \dfrac{1}{2} \times 0.5 \times 2^2 \[0.5em] = 1 \text{ J} \[0.5em]=21​×mass×velocity2=21​×0.5×22=21​×0.5×22=1 J∴ Potential energy = Kinetic energy = 1.0 J

Question 19

Question 6A body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity increases from u to v. How much work is being done by the force?

Accepted Answer

Mass = mChange in velocity from u to v, so acceleration acts (a)Force = fWork done by the force = force × displacementFrom equation of motion we know,v2=u2+2aS∴S=v2−u22av^2 = u^2 + 2aS \[0.5em] \therefore S = \dfrac{v^2 -u^2}{2a}v2=u2+2aS∴S=2av2−u2​Substituting it in the formula for work done:W=m×a×(v2−u22a)W=(12×m×v2)−(12×m×u2)W=(12×m×v2)−(12×m×u2)W=12m(v2−u2)W=KEf−KEiW = m \times a \times (\dfrac{v^2-u^2}{2a}) \[0.5em] W = (\dfrac{1}{2} \times m \times v^2) - (\dfrac{1}{2} \times m \times u^2) \[0.5em] W = (\dfrac{1}{2} \times m \times v^2) - (\dfrac{1}{2} \times m \times u^2) \[0.5em] W = \dfrac{1}{2} m (v^2 - u^2) \[0.5em] W = KE_f - KE_iW=m×a×(2av2−u2​)W=(21​×m×v2)−(21​×m×u2)W=(21​×m×v2)−(21​×m×u2)W=21​m(v2−u2)W=KEf​−KEi​

Question 20

Question 1Show that the sum of kinetic energy and potential energy (i.e., total mechanical energy) is always conserved in the case of a freely falling body under gravity (with air resistance neglected) from a height h by finding it when(i) the body is at the top,(ii) the body has fallen a distance x,(iii) the body has reached the ground.

Accepted Answer

We know that,Kinetic energy + potential energy = constantSo, when a body falls from a height h under free fall(i) At the position A — height hInitial velocity = 0Kinetic energy K = 0Potential energy U = mghAs, Total energy = KE + PETotal energy = 0 + mghTotal energy = mgh       (1)(ii) At the position B — when it has fallen a distance x.Then, velocity at B = v1Then u = 0, s = x, a = gFrom equation of motion:v2 = u2 + 2aSv12 = 0 + 2gx = 2gx∴Kinetic energy K=12×m×v12=12×m×2gx=mgx\therefore \text{Kinetic energy K}= \dfrac{1}{2} \times m \times v_1^2 \[0.5em] = \dfrac{1}{2} \times m \times 2gx \[0.5em] = mgx∴Kinetic energy K=21​×m×v12​=21​×m×2gx=mgx∴ Potential energy U = mg (h – x)Hence, total energy = K + U = mgx + mg (h – x) = mghTotal energy = mgh       (2)(iii) At position C (on the ground) —Let the velocity acquired by the body on reaching the ground be v2.Then u = 0, s = h, a = gWe know,v2= u2 + 2aSv22 = 0 + 2ghv22 = 2gh∴Kinetic energy K=12×m×v22=12×m×2gh=mgh\therefore \text{Kinetic energy K}= \dfrac{1}{2} \times m \times v_2^2 \[0.5em] = \dfrac{1}{2} \times m \times 2gh \[0.5em] = mgh∴Kinetic energy K=21​×m×v22​=21​×m×2gh=mghAnd potential energy U = 0 (at the ground when h = 0)So, total energy = K + U = mgh + 0Total energy = mgh       (3)Thus, from equation (1), (2) and (3) we note that the total mechanical energy i.e. the sum of kinetic energy and potential energy always remain constant at each point of motion and it is equal to the initial potential energy at height h.

Work, Energy and Power

Exercise 2A Multiple Choice Type

Exercise 2A Numericals

Exercise 2B Multiple Choice Type

Exercise 2B Numericals

Exercise 2B Short Questions

Exercise 2C Long Questions

Exercise 2C Numericals