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# Tag Archives: speed

## Time Travel Again !

Davide Writes: (The email is posted as such. SO, sorry for the telegraphic language and poor spelling and grammar)

“Hello there i have one more question,a crazy one:) according to Einstein we need a enormous amount of energy to push matter at a speed of light,now for me the real problem there is that an atoms at a speed of light will break a part right?so even if we will find a way to push it,we will still failed.I was wander are we looking to the right way?is it photon, so the world of small with no mass only work? now here my crazy question,lets speculate a bit,we have two red dwarf pretty close to earth Barnard’s Star” only “5.9 ly and Wolf 359 7.7ly a baby step compere the universe,if we can send a same kind of mission the we have done with Stardust Nasa’s comet sample and collect a piece the size of a soccer ball back to earth,we could work on it outside earth atmosphere in the space base.Since we can’t go in the first place at a speed of light let’s says that the all deal will last 150 year to go and back,(assuming that we use a new energy to go bit faster)in the mean time the space base will be pretty developed and maybe we even found a new power to push to the speed of light(i remind you that we are speculate)we could put our hands on the sample using electromagnetism so that we avoid any risks and then send it at a seed of light.What i try to say is that a single atom is weak but billions of it together is a different story,here an example, lets imagine to make a spherical shape out of Lego(the toy) and send it in to space at a great speed,it will break pretty soon,now we make another one but we glue the piece together,it will last bit longer,again another with glue and a put net around it will last even longer so instead of a Lego we use our sample,the glue in the inside is the immense mass that will have (piece of red dwarf) net outside the immense gravitational pull ,now do you think that it will be enough to keep it together at a speed of light?regardless the energy that we need to push it? after all even in the past we believed that going faster than a speed of sound a plane will brake a part or something bad could happen,instead only a big BUM wen we reach it and nothing more,so maybe if we can go faster than light nothing will happen,or maybe we find a new way to travel,we can even give a name for fun “a speed of matter” what do you think about it?i know it sound like sci fi but hey physics it self is blowing mind science right? Thank you kindly for your time:)”

**I don’t like to respond now.**

** Responses from visitors invited**

## Problem from Kinematics

The speed of a train changes from36km/h to 72km/h in 10sec. calculate the distance travelled during this time

**Answer**:

u=36km/h = 10 m/s

v=72 km/h=20 m/s

t=10 s

S=?

use **v = u+ at** to find a

a = 1 m/s^{2}

Use **v ^{2}-u^{2}=2aS** to find S

S = (400-100)/2 = **150 m**

## Ron writes on Light and Relativity

Let me start by saying I am NOT a physicist. That much should be painfully clear soon enough. Please bear with me as I am not certain I have the language and in depth **knowledge** to explain myself adequately.

A few years ago I began experiencing parallels between the world of human behavior and **physics**. At the **time** it seemed simple enough. That is until I started teaching myself **physics**. But at its core it seemed to make sense, the rules of life recycle themselves in different forms. Out of simple ideas comes complexity. Newton’s laws, thermodynamics… pascals principle, snells law, red and blue shift, wave **particle** duality and on and on… all seem to have their parallels in human behavior and on the surface seem to follow the same mathematical equations. Then I ran into the most famous of **physics** equations E=MC^2. It makes sense in human behavior. The **energy** (work) we can get out of an individual is relative to that individuals’ **mass**. The more **matter** we attach to the individual, **knowledge**, life experience etc (therefore increasing their **mass**) the more we can get out of them in the form of **energy**… but then the problem of the **speed** of **light** squared. The only known constant (**light** **speed**) is a problem in human behavior. At least to this point I know of no known behavioral constant.

I juggled the **idea** of it being a relative constant. Constant for the individual but relative as it would differ for everyone. The **speed** of cognition, or thought **speed**, would remain the same in potential throughout life of a given individual. Or that it was just a theoretical potential that humanity had yet to obtain. I even juggled the **idea** of it being a collective ability, but all of these options cause problems with the original equation. This has forced me to contemplate the nature of **light** and left me with questions I simply lack the understanding of **physics** to answer.

Is the **speed** of **light** truly constant or is it only constant as it relates to the big picture? As in: our **perception** of **time** as it relates to all **time** that has ever existed would appear as a single point in **time**. The older we get the more **time** seems to “fly by” if we as humans could continue to live for 13 billion years would the **perception** of an hour become so perceptively small that we wouldn’t even know it has passed? Is it therefore possible we can only understand and therefore measure the **speed** of **light** at a specific point in **time**, even if we try to come back and remeasure and compare the **speed** of **light** now with the **speed** of **light** fifty years from now the difference between the two would be imperceptivity small as a result of the displacement in **time** as it relates to the whole of **time**?

If the four dimensionality of **time**/**space** is linked shouldn’t **time** expand as **space** expands? And vice versa… if the **speed** of **light** is to remain constant as measured under such conditions isn’t **light** actually slowing down/speeding up over **time** as it relates to the whole of existing **time**/**space**? It’s a **distance** displacement problem…. If points A and B are actually farther apart but **light** travels the same **distance** in the same **time** the “**speed**” may seem constant but **time** has actually expanded to give the **light** more “**time**” to cover that **distance**. The **speed** of **light** would therefore be constant as a **relation** to **perception** and not as it relates to physical principles. Like the fact that the **perception** of **time** changes as we get older even though the actual measure of **time** remains the same.

How can **light** exist forever? At the **speed** of **light** we theoretically freeze **time** for that **photon** but that would also require an **infinite** amount of **energy** to obtain and maintain. So even a subatomic **particle** with a lifespan of a nanosecond would appear to exist for all eternity but in actuality would still only exist for a nanosecond. Because quite simply…it can’t have **infinite** **energy** and if it can I don’t understand how. Thermodynamics: no system is a perfect system and will experience **energy** loss, Newton: equal and opposite reaction, if something begins it has to end to balance the equation.

Is it therefore possible that **light** is born of the fourth dimension… we experience it in the dimensionality of **space** as long as it loses its **energy** to the three **dimensions**. A **photon** folds and pushes its way through **space**/**time** The initial **energy** of the **photon** is high and generates bigger leading waves which resist the **photon** holding it from passing the “**speed** of **light**” as the **energy** dissipates/**photon** begins to die those folds restricts the **photon** less allowing it to maintain the **speed** of **light**. We experience **light** because of its “ripples” in three-**dimensions** **light** dissipates as the waves of **space** become less folded in front of the **photon**. But this would mean the **photon** eventually loses enough **energy** that it can no longer be perceived in the third dimension…. So what happens to it?? Imagine : A man running through a corn **field** has to exert the **energy** to push the stalks aside but over **time** if the corn stalks are slowly spaced out even as he loses **energy** he can maintain his pace, because there is less impeding his path. Our **perception** of **light** would be like being in a helicopter looking down on the **field**. In the begging the **field** is densely packed with cornstalks and they slowly spread out until there is none. We only know the man is there while he is running through the corn because he pushes the stalks aside and we see that movement but once there are no stalks or he lacks the **energy** to continue running we no longer have a way of measuring his presence.

That brings me for some reason to a theory of Dark **matter**. Why? Well what happens to **photons** that no longer move with enough force to be visible. Like that guy running from a cornfield into an empty **field**. How would we know he is still there? If it was an **infinite** number of guys all stopping in that empty **field** we would know they are there by the depression their weight leaves in the **field**. Or more accurately by the stones in the **field** rolling towards a depression we cant actually see. The thought: An **infinite** number of “massless” (or perceived **mass** less) subatomic particles would still have **infinite** **mass**. Infinite **mass** would supply more then enough gravitational pull even spread out over **infinite** **distance** to cause the continued and speeding expansion of **space** as more and more visible **photons** “die” contributing its “dark masslessness”. Any dark **matter** existing within the universe would act as force… a moving invisible **mass** existing only on the 4th dimension pushing upon any objects in its way. But because **energy** propagated internally expanding outward in all **dimensions** would compound on the outside as those **energy**’s converge.

I hope you followed those questions. I don’t know what these ideas would do to theoretical **physics** but it would allow for the relativity of cognitive **speed** between individuals and reopen the door to “what the hell is dark **matter** in **relation** to human behavior” but that’s something different all together.

## 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)**.

## Potential

Devna asked.

find the potential at A ….a prson brings a mass of 1 kg from infinity to the pt A .initially the mass was at rest but it moves at a speed of 2 m/sas it reches A .the work done by person on mass is -3 J

## A Few Problems from Kinematics

- A body falling freely from rest has velocity v after it falls a height h . Calculate the distance it should fall down further for its velocity to become double
- A particle in uniform acceleration in a straight line has a speed of v m/s at position x meter is given by √(25-16x). What is the acceleration of the particle

## Average Speed

Naman Asks:

“A car travels from A to B with a speed of 30km/h and returns back to A with a speed of 50km/h.Find distance,displacement,average speed and average velocity ifdistance between A and B is 100km.”

Answer:

Distance 200 m

Displacement 0

Average velocity 0

Average speed = = 37.5 kmph