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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?
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.
In what sequence should these chapters be prepared for my exams:-
units and dimensions,kinematics,laws of motion,work power and energy,motion of system of particles and rigid body dynamics,gravitation.
Apoorv Katoch asked
It is better to follow the same pattern as given in NCERT text book
“We know that sound can cause some mechanical work(eg:when we keep our hand in front of a large speaker,we feel something hitting the back of our hand), but why this doesn’t happen with light??” -Manishankar asks.
Nischal Singh Dangol Asked:
How to get A in physics?
Whatever the syllabus or course, an A grade means “Excellent work indicating a clear mastery of the subject material”
This requires evidently a basic understanding of the subject, thorough with the fundamentals and the ability to apply the principles to solve problems creatively. One should be able to make suitable connections between concepts logically to solve the problem in hand. (Remember, Physics is often referred to as a Problem Solving Discipline)
A grade in Physics comes naturally if you start loving the subject and take active interest in applying the concepts of Physics to solve problems (in subject as well as in day to day life situations)
Also important :
- Maintaining good rapport is also important in getting an A grade.
- Attend all lectures/classes and never bunk
- Do your assignments sincerely and submit in time
Wish you all the best.
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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.
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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
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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.
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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.
Akshay posted a numerical:
1. A disk of mass m=50g slides with zero initial velocity down an incline plane set at an angle 30 degrees to the horizontal. Having traversed the distance of 50 cm along the horizontal plane, the disc stops. Find the work performed by friction forces over the whole distance. Assume friction coefficient is 0.15 for both horizontal and incline planes.
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