LUIT Valley Academy : Jorhat, Class XI Project
Introducing “The FORWARDERS”
MEMBERS:
DEBOJANEE DUTTA
DIPSIKHA SAIKIA
PRATIKSHA SHARMA
PRIYADARSHNI BORDOLOI
ANUSHUYA GOGOI
SANGHAMITRA GOGOIUnder the Guidance of
Dr. Sanjan Hazarika (Prof. in Charge)
Relation:The work done by a force is measured by a
dot product of force and displacement.If a force F acting on a body displaces it through a displacement d, then the work done by the force is given by W = F.d
If the force is not along the direction of displacement, the component of force along the direction of displacement should be considered
UNITS OF WORK The S.I. unit of work is joule
1J = 1N × 1m = 1Nm
The CGS unit of work is erg.
1erg = 1 dyne× 1cm = 1dyne cm
1J = 10⁷ erg
TYPES OF WORK: Positive work:
Work done(W)=F.Scosθ=positive value[0≤θ≤90⁰]
Negative work:
Work done(W) = F.Scosθ = negative value [90⁰≤θ≤180⁰]
Energy can neither be created nor bedestroyed
It is transferred from one object to another or from one form to another form
Law of conservation of energy
Momentum P = mv It is the amount of motion contained in a body
P a m
a v
KE = ½ mv2 mm
vm
2
P
2
222
2mE P
CONSERVATION OF
MECHANICAL ENERGY
This principle states that if only the conservative forces are doing work on a body , then its mechanical energy (KE+PE) remains constant.
As the force is conservative , then change in potential energy is given by-
ΔU= -F(x)Δx = negative of the work done
CONSERVATIVE FORCE NON-CONSERVATIVE FORCE
If the work done by the force is displacing an object depends onlt on the initial and final positions followed between the initial and final positions,such a force is known as conservative force
E.g: gravitational force
If work done by aforce in displacing an object from one position to anotherdependsupon the path between the two positions, such a force is called non-conservative force.
E.g:force of friction,tension
Rotational and translational KEKE of translation = ½ mv2
KE of rotatation = 1/2Iw2
Total Energy = ½ mv2 +1/2Iw2
= ½ mv2 +1/2(2/5mR2)v2/R2
= ½ mv2 + 1/5 mv2
= 7/10 mv2
PROBLEMS:1. A body of mass 5 kg initially at rest is subjected to a force of
20N. What is the kinetic energy acquired by the body at the end of 10 seconds?
2. If the kinetic energy of a body increases by 300%, by what % will the linear momentum increases?
Sol 1. a = F/m = 20/5 = 4m/s2
u = 0
v = u +at = 0 + 4 . 10 = 40 m/s
KE = ½ mv2 = ½.5.40.40 = 4000J
Sol 2.
..100..100P
P increasePercent
P P - 2P P -P P
momentumin Increase
2P
2mE2 2m(4E)
4E E100
300 E E E2m P
2mE P
cpcpx
POWER: Power of a person or machine is defined as the rate at which work is done or energy is transferred. If a person does work W in time t, then its average
power is given by-
Average power(Pav) = rate of doing work
=work done/time taken
=W/t
PROBLEMS: A man weighing 60 kg climbs up a staircase carrying
a load of 20 kg on his head. The staircase has 20 steps each of height 0.2 m. If he takes 10 seconds to climb, find his power.
The human heart discharges 75 ml of blood at each beat against a pressure of 0.1 m of Hg. Calculate the power of heart assuming that pulse frequency is 80 beats per minute. Density of Hg = 13.6×10³ kgm¯³.