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# Tag Archives: kinetic energy

## Work done and Kinetic/Potential Energy

Hi!

My Physics Class 11 question says that a climber who is 65kg climbs 600m. What is the work done by the climber?

I thought I would have to add both the change in KE and PE. But the answer says that you should simply use mgh (PE).

This is confusing because there should be some work done through movement. Why is KE omitted from calculating the work done that in this case?

Leo asked

**Answer:**

Many problems in Physics are solved by making certain approximations and assumptions to avoid complications. If we consider all factors contributing towards the expenditure of energy there may still different factors, but of less importance for the problem under consideration.

In the question we assume that there is no **change in Kinetic Energy **involved. Work is not to be done to maintain the KE but to change the KE. Though there may be some changes in speed during the process, we disregard the changes and concentrate only on the change in potential energy, which is actually the work done against gravity.

Even if there is a change in Kinetic Energy, then also the work done against gravity will only be the change in potential energy (**mgh**). He may be doing work against friction and also the change his speed during the climbing process, but the work done against gravity would remain the same (mgh)

If the question had mentioned this (work done against gravitation) then confusion could have been avoided.

Hope that the idea is clear now. If you need further clarifications please respond via the comment form.

## Where is the light energy going?

where is the sun’s light enery going once the sun has set?

or

where is the light going when we switch off the torch facing an object. if light gets reflected how long it will do so. and if it gets absorbed, then will it add to the potential energy of the object that is facing the torch?

or

Is there any cycle (like hydrocycle)to prove that energy can neither be created nor be destroyed?

Asked Dingu Sagar

## WAVE MOTION QUESTIONS AND ANSWERS FOR CLASS XI

**One mark questions with answers**

**Q1.****body** of **mass** 2kg. is at restand is hit by a **mass** of 4kg. moving with 3m/s, find fraction of the momentumretained by the moving **body** assuming the collision to be elastic and head-on.

**Ans1.**_{2}/m_{1}

(n – 1)/(1 + n) is the fraction of the momentum retained by the moving body so

(n – 1)/(1 + n) = (2 – 1)/(1 + 2) = 1/3.

**Q2.****mass** of the moving **body** ismuch greater than the **mass** of the **body** at **rest** than what is the approximatevilocity of the moving **body** after head-on collision?

**Ans2. ****mass** is muchgreater than the **mass** at **rest** then after the collision the heavier **mass** keepson moving with the same **velocity** and in the same direction.

**Q3.****point** the potentialenergy of a **body** is taken to be zero?

**Ans3.****energy** of the bodyat the surface of the earth is taken to be zero (Potential **energy** = mgh, whereh is the **height** of the **body** from the surface.).

**Q4.****work** done on a **body** by aforce depend upon the path followed by it?

**Ans4. ****force** isconservative then it does not depend but if it is non-conservative (friction)then it depends.

**Q5.****body** hits the **ground** from aheight h_{1} and rebounds to a **height** h_{2} after havinginelastic collision with the **ground** then what is the coefficient ofrestitution?

**Ans5. **_{2}/h_{1})

**Q6.****body** hits the **ground** with 50m/s **velocity** and has inelastic collision with the **ground** then with whatvelocity it will rebound if the coefficient of restitution is 0.2.

**Ans6. **_{2}/v_{1}where v_{1} is the **velocity** with which the **body** hits the **ground** and v_{2}is the **velocity** of rebound.

e = 0.2 = v_{2}/50, so v_{2} = 10 m/sec.

**Q7.****body** at **rest** explodes in threefragments. Is it possible that two equal parts move in mutually perpendiculardirections with the same **velocity** and third **mass** moves midway between the two?

**Ans7. **

**Q8.****speed** of a moving vehicleis increased by 200% then how much should be the change in the retarding forceto stop the vehicle over half the previous distance?

**Ans8. **^{2} = F.S where Fis the retarding **force** and S is the distance over which the vehicle comes torest. When ‘v’ is increased by 200% then K.E. increases by 800%. As S is halvedthen F should be made 16 times.

**Q9.****work** is done in compressinga spring from 0 cm to 6 cm then find the **work** done in compressing the same from3cm to 6 cm.

**Ans9.**_{2}^{2}– x_{1}^{2}). This is the **work** done when the spring iscompressed from x_{1} to x_{2}.

20 = (1/2)K (36 – 0)

W = (1/2)K (36 – 9)

Solving the above equations we get W = 15 J.

**Two mark questions with answers**

**Q1.****ball** is dropped from **rest** at aheight of 20m. If it loses 30% of its kinetic **energy** on striking the **ground**,what is the **height** to which it bounces? How do you account for this loss inkinetic **energy**?

**Ans1.****ball** acquires avelocity ‘v’ after falling through a **height** of 20m.

Because the ball is dropped from rest, hence u = 0.

Hence, v^{2} = u^{2} + 2as

= 0 + (2 ** x** 10

**20) = 400**

*x*So, v = 20 m/s

Kinetic energy of the

**ball**just before hiting the ground

= (1/2)mv

^{2}= (1/2)m(400) = 200m Joule

Because the

**ball**loses 30% of the kinetic

**energy**on striking the

**ground**, hencekinetic

**energy**retained by the

**ball**after striking the

**ground**= 70% of 200m J

= 140m J

The

**energy**loss is due to the inelastic collision with the

**ground**.

**Q2.****energy** being consumedin planetary motion.

**Ans2.****body** whichrevolves round the star (the sun). The **force** which is responsible for circularmotion, is called centripetal **force**. The direction of the centripetal **force** isalways towards the centre. Thus, the angle between **force** F and displacement Sis q^{o} at every **point**.Work done in moving planet, W = **F.S** = FSCos^{o}.So, W = 0. Hence,no energyis being consumed in planetary motion.

height=82src=”./xi%20work,%20power%20and%20energy_files/image001.gif”alt=”workpowerandenergyf06q16i.gif (1336 bytes)” v:shapes=”_x0000_i1027″>

**Q3.****energy**?

**Ans3.****energy** (Power).The potential **energy** of water storedat a **height** is converted into K.E. when water is made to rush down. This fallof water is used to rotate the turbine and the coil and armature of generatoris rotated and electricity is produced . Thus, the K.E. of the fall of water isconverted into electrical form of **energy**. Hence the hydroelectric power-plantis an example of law of conservation of **energy**.

**Q4.****body** is dropped from aheight of 40m then after 3 inelastic collisions with the **ground** to which heightthe **body** will rise? (given: Coefficient of restitution = 0.5)

**Ans4.****body** is dropped from aheight of ‘H’ and ‘e’ is the coefficient of restitution then after ‘n’inelastic collisions with the **ground** the **body** rises to a **height** ‘h’ given by

h = H.e^{2n}.

h = 40 ** x** (1/2)

^{2 x 3}= 40

**(1/2**

*x*^{6}) = 40/64 = 0.625 m

**Q5.**

**Ans5.**_{1}v_{1}^{2}= (1/2)m_{2}v_{2}^{2} ………….(*i*)

If force of brakes be the same then m_{1}a_{1} = m_{2}a_{2}……….(*ii*)

If truck stops over a distance S_{1} then v_{1}^{2} =2a_{1}S_{1} ……..(*iii*)

If car stops over a distance S_{2} then v_{2}^{2} = 2a_{2}S_{2}………(*iv*)

From (*i*) and (*ii*)

(1/2)m_{1}v_{1}^{2} = (1/2)m_{2}v_{2}^{2}………..(*v*)

From (ii) and (v)

v_{1}^{2}/a_{1} = v_{2}^{2}/a_{2}…………(*vi*)

From (*iii*), (*iv*) and (*vi*)

2a_{1}S_{1}/a_{1} = 2a_{2}S_{2}/a_{2}.

S_{1} = S_{2}

Hence distances covered S_{1} and S_{2} are equal.

**Q6.****kg**/minute can it raise a **height** of 20m? (g = 10 m/s^{2})

**Ans6.**

If mass of water raised in one second = m kg.

Total work done in lifting water,W = mgh

Power P = W/t, but t = 1 minute = 60 sec.

4000 = mgh/60

4000 = (m ** x** 10

**20)/60**

*x*m = 1200

**kg**.

**Q7. ****length** 3m is suspendedvertically from a fixed **point**. It is given an angular displacement of 60^{o}in the vertical plane. If its **mass** per unit **length** is 2 **kg** then find the workdone?

**Ans7. ****mass** of the rod and’l’ be its **length** then

m = 2 ** x** 3 = 6

**kg**

If the rod is displaced through an angle qthen the

**work**done on it, W = mg(l/2)(1 – Cosq).

The effective length of the rod is taken to be (l/2) because in uniformdistribution of

**mass**the centre of

**mass**is at the geometric centre so

W = 6

**10**

*x***(3/2)(1 – Cos60) =45 J.**

*x*

**Three mark questions with answers**

**Q1.****mass** m* _{2}*is at

**rest**and

**mass**m

_{1}moving with

**velocity**u

_{1}hits itelastically, show that the fraction of the momentum transferred to the

**mass**atrest is 2n/(1 + n) where n is ratio of the masses.

**Ans1.**_{1} = (m_{1} -m_{2})u_{1}/(m_{1} + m_{2}).

v_{2} = 2m_{1}u_{1}/(m_{1} + m_{2}).

This is from the theory of conservation of momentum.

Momentum of the **mass** m_{2} after collision,

P_{2} = m_{2}v_{2} = (2m_{1}m_{2}u_{1})/(m_{1}+ m_{2})

Fraction of momentum transferred to m_{2}.

= (2m_{1}m_{2}u_{1})/(m_{1} + m_{2})m_{1}u_{1}= 2m_{2}/(m_{1} + m_{2})

= 2n/(1 + n) ….[because m_{2}/m_{1} = n]

**Q2.****work** to bepositive negative or zero? "Explain with example.

**Ans2. ****force** applied F and the displacement S movedby **body** *i.e.*

W = F.S,

W = FS Cos q

If q**work** is maximum.

It remains +ve for the angle q between 0^{o}to 90^{o}

height=45src=”./xi%20work,%20power%20and%20energy_files/image002.gif”alt=”tutor2phyworkpowerandenergyf06q2i1.gif (1062 bytes)” v:shapes=”_x0000_i1037″>**or**, q^{o} and360^{o} *i.e.*, if the displacement is in a directionoppisite to which the **force** is applied.

height=59src=”./xi%20work,%20power%20and%20energy_files/image003.gif”alt=”tutor2phyworkpowerandenergyf06q2i2.gif (1097 bytes)” v:shapes=”_x0000_i1038″>

Thus **work** is +ve if Cos q**work** done will be -veif Cosq*i.e.* ^{o} to 270^{o}.

If q^{o} then Cos 90^{o}= 0.

Hence **work** done W = FS Cos 90^{o} = 0.

Thus, W may be +ve, -ve or zero

**Q3.****length** of the pendulum is 2m, whatis the **speed** with which the bob arrives at the lowermost **point**? Given that itdissipates 10% of its initial **energy** against air resistance?

**Ans3.****energy** atthe highest position

= mg ** x** 2 Joules = 2mg Joules

Kinetic

**energy**at lowest position

= Potential

**energy**at the highest position – the

**energy**dissipitated againstair resistance or friction

= [mg

**2 – (10/100)**

*x***mg**

*x***2] Joule**

*x*= mg

**18/10 J**

*x*^{2}= mg

**18/10**

*x*or, v = 1.9 ms

^{-1}.

**Q4.****work**. Give their relations also.

**Ans4. **

W = FS**(1)** In S.I system,

If F = 1 **kg** weight or 1 **kg** **force** and S = 1m then,

W = (1 **kg** wt)(1m) = 1 **kg** m ……………(*i*).

Hence, one kgm is the gravitational unit of **work** in S.I (M.K.S) system and isdefined as the amount of **work** done if 1 **kg** **force** displaces a **body** through 1m inthe direction of the applied **force**.**(2)** In C.G.S system,

F = 1 gmwt and S = 1cm,

W = (1 gm wt) (1 cm) = 1 gm cm ………………(*ii*).

Hence, one gm cm is the gravitational unit of **work** and is defined as the amountof **work** done, if 1 gm **force** displaces a **body** through 1 cm in the direction ofthe applied **force**.

1 gm cm = 980 ergs.**NOTE**: 1 **kg** m = 9.8 Joules.

**Q5.**

**Ans5. **

One electron volt is the **energy** acquired by one elctron in moving it betweentwo **point** having a P.D of 1V.

Thus, 1eV = (1.6 ** x** 10

^{-19}) C

**1J/C = 1.6**

*x***10**

*x*^{-19}Joules.

**NOTE:**The other practical units used are

1 Million electron volt = 1 MeV = 10

^{6}eV, 1 MeV = 10

^{6}

**1.6**

*x***10**

*x*^{-19}J, 1 MeV = 1.6

**10**

*x*^{-13}Joules and

1 Billion eV = 10

^{9}eV, 1 BeV = 1.6

**10**

*x*^{-10}joules.

**Q6.****velocity** changes by 5%, find the change in the power of water?

**Ans6.**** x**Velocity = Rate of change of momentum

*x***velocity**={(

**mass**/time)

*x***velocity**}

*x***velocity**= {(adv)

**v}**

*x***v =adv**

*x*^{3}where ‘a’ is area of cross section, ‘d’ is the density of waterand ‘v’ is the

**velocity**of flow of water.

Therefore, Power of water is directly proportional to the cube of velocity ofwater so let P = Kv

^{3}(k is a constant and is equal to ‘ad’.)Taking log on both sides

log P = 3log v + log k

Differentiating on both sides

D

percentage change in power, DP/P

**100 = 3**

*x***5%**

*x*= 15%.

**Q7.****energy** of rushing outwater from a dam is used in rotating a turbine. The pipe through which water isrushing is 2.4 meters and its **speed** is 12 m/sec. Assuming that whole of kineticenergy of the water is used in rotating the turbine, calculate the currentproduced if efficiency of the dynamo is 60% and the station transmits power at240 kV. Density of water = 10^{3} **kg**/m^{3}.

**Ans7.**

r = radius of pipe = 1.2m, average speed of water v = 12 m/s

V = 240 kV = 240 ** x** 10

^{3}volt, density ofwater p = 10

^{3}

**kg**/m

^{3}.

Now, kinetic

**energy**of rushing water per second i.e.

Power P = (1/2)(

**mass**flowing per sec)

**v**

*x*^{2}

= (1/2)p

^{2}(l/t)

^{2}

= (1/2)p

^{2}

^{3}

= (1/2)

**3.14**

*x***(1.2)**

*x*^{2}

**10**

*x*^{3}

**(12)**

*x*^{3}watt

= 3.9

**10**

*x*^{6}watt

current = output power/voltage

= (60% of power P)/(240

**1000)**

*x*= [(60/100)

**3.9**

*x***10**

*x*^{6}]/(240

**1000) = 9.75 amp.**

*x*

**Five mark questions with answers**

**Q1.****mass** m isaccelerated from **rest** when a constant power P is supplied by its engine; showthat :

(a) The **velocity** is given as a function of time by

v = (2Pt/m)^{1/2}

(b) The position is given as a function of time by

s = (8P/9m)^{1/2}t^{3/2}.

(c) What is the shape of the graph between **velocity** and **mass** of the vehicle ifother factors remain same?

(d) What is the shape of the graph between displacement and power?

**Ans1.***i.e.*, m ** x** (dv/dt)

**v =P [as F = ma = m**

*x***(dv/dt)]**

*x*After rearranging and integrating on both sides

ò

**dt**

*x*(v

^{2}/2) = (P/m)

**t + C**

*x*_{1}

Now as initially the

**body**is at

**rest**,

*i.e.*, v = 0 at t = 0, so C

_{1}= 0.

v = (2Pt/m)

^{1/2}…………(1)

(b) By definition v = (ds/dt),

Using eq (1) above,

ds/dt = (2Pt/m)

^{1/2}

On integration we get

ò

^{1/2 }dt

s = (2P/m)

^{1/2}

**(2/3)**

*x***t**

*x*^{3/2}+ C

_{2}.

Now, as at t = 0, s = 0, so, C

_{2}= 0

s = (8P/9m)

^{1/2}t

^{3/2}.

(c)

height=104src=”./xi%20work,%20power%20and%20energy_files/image004.gif”alt=”5a1.gif (1325 bytes)” v:shapes=”_x0000_i1046″>

(d)

height=102src=”./xi%20work,%20power%20and%20energy_files/image005.gif”alt=”5a1i.gif (1265 bytes)” v:shapes=”_x0000_i1047″>

**Q2.****mass** m and lengthl swings back and forth up to a maximum angle q_{0} with the vertical. When at an angle q, what is its (a) potential **energy**, (b) kinetic **energy**, (c) **speed**, and(d) tension?

**Ans2.**

height=149src=”./xi%20work,%20power%20and%20energy_files/image006.gif”alt=”5a2.gif (2041 bytes)” v:shapes=”_x0000_i1048″>

Taking the reference level at the lowest point R, we have

h_{P} = l – l cos q_{0}_{0}

h_{Q} = l – l cos q = l(1 – cos

So (a) potential **energy** at Q relative to R will be

PE = mgh_{Q}

PE = mgl(1 – cos q

(b) PE at P = mgh_{P} = mgl(1 – cos q_{0})

KE at P = 1/2 ** x** mv

^{2}= 0

so, total mechanical

**energy**at P = mgl(1 – cos q

_{0}) …….(

*i*)

Now, if K

_{Q}is the KE at Q,

then using eq. (

*i*)

mechanical

**energy**at Q = K

_{Q}+ mgl(1 – cos

*ii*)

But by conservation of mechanical

**energy**between P and Q

K

_{Q}+ mgl(1 – cos q) = mgl(1 -cos q

_{0}

*i.e.*, K

_{Q}= mgl(cos q– cos q

_{0}

(c) If v is the

**speed**at

**point**Q, from eq. (b)

1/2

**mv**

*x*^{2}= mgl(cos

_{0})

*i.e.*, v = height=17src=”./xi%20work,%20power%20and%20energy_files/image007.gif”alt=”i2.gif (1049 bytes)” v:shapes=”_x0000_i1049″>.

(d) If ‘E’ is the

**energy**at Ðq, then itis equal to mgl(1 – Cosq

^{2}.

Since the

**energy**remains constant throughout, E = E

_{o}.

mgl(1 – Cosq

^{2}= mgl(1 -Cosq

_{o}

or mv

^{2}= 2mgl(Cosq – Cos

_{o}

Therefore, tension ‘T’ at q would begiven by

T = mv

^{2}/l + mgCosq = mg Cos

_{o}

or T = 3mgCosq

_{o}

**Q3.**What do youmean by **work** in the language of physics? Give its absolute and gravitationalunits. Give two illustrations of zero **work**, negative **work** and positive **work**.

** Ans.(Try yourself)**.

**Q4.**How will youfind **work** done by a variable **force** mathematically and graphically?

** Ans.(Try yourself)**.

**Q5.**What do youmean by conservative and non-conservative forces? Give their importantproperties.

** Ans.(Try yourself)**.

**Q6.**What do youmean by gravitational potential **energy**? Show that gravitational potentialenergy is independent of the path followed.

** Ans.(Try yourself)**.

**Q7.**If a **body** iskept on the top of a rough inclined plane, find the expression for

(i) work done in bringing it down to the bottom of the plane with constantvelocity

(ii) **work** done in moving it up the plane with constant acceleration

(iii) **work** done in moving it down the plane with constant acceleration.

** Ans.(Try yourself)**.

## Conservation of energy Problem

A **ball** is dropped from a **height** of **10m**. If the kinetic **energy** of the **ball** reduces by 40% after striking the **ground**, how high can the **ball** bounce back? (**Elsy James** asked)

## Momentum and Kinetic Energy

Two bodies one light and other heavy have equal momentum.Which of them has higher kinetic energy?

(Aajma Manoj asked)

Answer:

The lighter body will have greater kinetic energy.

Explanatiom

Momentum, p=mv

squaring both sides, p^{2} = m^{2}v^{2}

Kinetic Energy = 1/2 mv^{2} = p^{2}/2m

Therefore, for bodies of equal momenta, kinetic energy is inversely proportional to mass.

## Conservation of Angular Momentum and Kinetic Energy

Imagine a rod 2 meters long fixed at its center, 2 identical masses are placed 0.5 meters away on opposite sides , as the rod spins they slide towards the edges , according to the formula k.E.=0.5*I*ω^{2} the kinetic energy decreases , but since its a closed system why does the kinetic energy decrease ?

**Answer**:

The decrease in KE accounted by the work done in displacing the masses

## A problem from Transformation of Energy

Rosie asks:

“I can’t figure out how to solve this problem:

The total mass of a motorcycle and rider is 250kg. During braking, they are brought to rest from 16m/s in a time of 10s. What is the maximum amount of energy converted into heat by the brakes?

I don’t know whether to use WD = F x d or another set of formula?”

Answer:

This question is based on conservation of energy. The heat energy liberated is equal to the change in kinetic energy. Since the bike is brought to rest, the change in KE is equal to its initial KE.

The answer assumes that the entire KE is converted into heat energy alone.

Heat =Change in KE =