Home » Posts tagged 'kg'
Tag Archives: kg
One mark questions with answers
(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.
e = 0.2 = v2/50, so v2 = 10 m/sec.
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
Because the ball is dropped from rest, hence u = 0.
Hence, v2 = u2 + 2as
= 0 + (2 x 10 x 20) = 400
So, v = 20 m/s
Kinetic energy of the ball just before hiting the ground
= (1/2)mv2 = (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.
height=82src=”./xi%20work,%20power%20and%20energy_files/image001.gif”alt=”workpowerandenergyf06q16i.gif (1336 bytes)” v:shapes=”_x0000_i1027″>
h = H.e2n.
h = 40 x (1/2)2 x 3= 40 x (1/26) = 40/64 = 0.625 m
If force of brakes be the same then m1a1 = m2a2……….(ii)
If truck stops over a distance S1 then v12 =2a1S1 ……..(iii)
If car stops over a distance S2 then v22 = 2a2S2………(iv)
From (i) and (ii)
(1/2)m1v12 = (1/2)m2v22………..(v)
From (ii) and (v)
v12/a1 = v22/a2…………(vi)
From (iii), (iv) and (vi)
2a1S1/a1 = 2a2S2/a2.
S1 = S2
Hence distances covered S1 and S2 are equal.
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 x 20)/60
m = 1200 kg.
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 x 10 x (3/2)(1 – Cos60) =45 J.
Three mark questions with answers
v2 = 2m1u1/(m1 + m2).
This is from the theory of conservation of momentum.
Momentum of the mass m2 after collision,
P2 = m2v2 = (2m1m2u1)/(m1+ m2)
Fraction of momentum transferred to m2.
= (2m1m2u1)/(m1 + m2)m1u1= 2m2/(m1 + m2)
= 2n/(1 + n) ….[because m2/m1 = n]
W = F.S,
W = FS Cos q
It remains +ve for the angle q between 0oto 90o
height=45src=”./xi%20work,%20power%20and%20energy_files/image002.gif”alt=”tutor2phyworkpowerandenergyf06q2i1.gif (1062 bytes)” v:shapes=”_x0000_i1037″>
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
Hence work done W = FS Cos 90o = 0.
Thus, W may be +ve, -ve or zero
= 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 x 2 – (10/100) x mg x2] Joule
= mg x 18/10 J
or, v = 1.9 ms-1.
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.
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 x1J/C = 1.6 x 10-19 Joules.
NOTE: The other practical units used are
1 Million electron volt = 1 MeV = 106 eV, 1 MeV = 106 x1.6 x 10-19 J, 1 MeV = 1.6 x10-13 Joules and
1 Billion eV = 109 eV, 1 BeV = 1.6 x 10-10joules.
Therefore, Power of water is directly proportional to the cube of velocity ofwater so let P = Kv3 (k is a constant and is equal to ‘ad’.)Taking log on both sides
log P = 3log v + log k
Differentiating on both sides
percentage change in power, DP/P x100 = 3 x 5%
r = radius of pipe = 1.2m, average speed of water v = 12 m/s
V = 240 kV = 240 x 103 volt, density ofwater p = 103 kg/m3.
Now, kinetic energy of rushing water per second i.e.
Power P = (1/2)(mass flowing per sec) x v2
= (1/2) x 3.14 x (1.2)2x 103 x (12)3watt
= 3.9 x 106 watt
current = output power/voltage
= (60% of power P)/(240 x 1000)
= [(60/100) x 3.9 x 106]/(240x 1000) = 9.75 amp.
Five mark questions with answers
(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)1/2t3/2.
(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?
i.e., m x (dv/dt) x v =P [as F = ma = m x (dv/dt)]
After rearranging and integrating on both sides
(v2/2) = (P/m) x t + C1
Now as initially the body is at rest, i.e., v = 0 at t = 0, so C1= 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
s = (2P/m)1/2 x (2/3) xt3/2 + C2.
Now, as at t = 0, s = 0, so, C2 = 0
s = (8P/9m)1/2 t3/2.
height=104src=”./xi%20work,%20power%20and%20energy_files/image004.gif”alt=”5a1.gif (1325 bytes)” v:shapes=”_x0000_i1046″>
height=102src=”./xi%20work,%20power%20and%20energy_files/image005.gif”alt=”5a1i.gif (1265 bytes)” v:shapes=”_x0000_i1047″>
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
hP = l – l cos q0
hQ = l – l cos q = l(1 – cos
So (a) potential energy at Q relative to R will be
PE = mghQ
PE = mgl(1 – cos q
(b) PE at P = mghP = mgl(1 – cos q0)
KE at P = 1/2 x mv2 = 0
so, total mechanical energy at P = mgl(1 – cos q0) …….(i)
Now, if KQ is the KE at Q,
then using eq. (i)
mechanical energy at Q = KQ + mgl(1 – cos
But by conservation of mechanical energy between P and Q
KQ + mgl(1 – cos q) = mgl(1 -cos q
i.e., KQ = mgl(cos q– cos q
(c) If v is the speed at point Q, from eq. (b)
1/2 x mv2 = mgl(cos
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
Since the energy remains constant throughout, E = Eo.
mgl(1 – Cosq
or mv2 = 2mgl(Cosq – Cos
Therefore, tension ‘T’ at q would begiven by
T = mv2/l + mgCosq = mg Cos
or T = 3mgCosq
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.
Q4.How will youfind work done by a variable force mathematically and graphically?
Q5.What do youmean by conservative and non-conservative forces? Give their importantproperties.
Q6.What do youmean by gravitational potential energy? Show that gravitational potentialenergy is independent of the path followed.
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.
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
Fusion gives more energy than fission right? When a uranium nucleus fissions into two daughter nuclei fragments, an energy of ~200 MeV is released. Whereas in a fusion of deuterium to give helium atom, only 18.3 MeV is released (which is the maximum?). Yes, the nucleus of uranium are heavier but the same deuterium fusion occurs in the stars and gives off a much larger amount of energy.
If we consider energy released per reaction (fission or fusion), energy released is more in the case of fission.
If we consider energy released per unit mass (say one kg Uranium and 1 kg hydrogen), the energy released in fusion is very much larger than in the case of fission.
In fusion about 4 amu releases approx 20 MeV whereas in fusion about 236 amu releases approx 200 MeV
a force of 20 N is applied parallel to the surface of a horizontal table is just sufficient to make a block of mass 4.kg move on the table calculate the coefficient of friction between the block and the table
It is correct to say the weight of the body is 10 kg? (Khalid asked)
As such it is not correct. But, kg. wt (kilogram weight) is the unit of weight and therefore someone simply says kg instead og kgwt hoping that it is understood that what they mean is kgwt.
How can we define one Newton? (Asked Pawas Rituja)
One newton is defined as that force which when acts on a body of mass 1 kg produces an acceleration of 1 m/s^2 in it
What is the weight of a body having a mass of 10 kg if it is falling freely?
Answer: The apparent weight of a freely falling body is zero.
When a body falls freely, it is falling with an acceleration =g. Thus the entire gravitational force acting on it is utilized in accelerating it and hence its weight is not felt.