work-energy-power (i) work :- work done by a force is the product of the force and the distance...

WORK-ENERGY-POWER

(i) WORK:-

Work done by a force is the product of the force and the distance moved by the point of application in the direction of the force. It is a scalar quantity.

Work done=F × cosα × s α= angle of inclination of force with the

direction of motion

s=displacement of force from A to B

Unit: Nm ( Joule )F F

s

α α A B

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(ii) POWER:-

It is defined as the time rate of doing work.

Power = work done /Time= (force × distance) /Time

= force × velocity

Unit: (Nm)/s [watt] (kN m)/s [kilo watt]

1 metric H.P=735.75 watts.

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iii) Energy:-

It is defined as the capacity to do work. It is a scalar quantity.

Unit :- N m (Joule)

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From Newton’s second law of motion∑ F =(W/g) × a -----------(1) Also a = dv/dt =( dv/ds) ×( ds/dt) = v × dv/ds

sub in (1) ∑ F=( W/g ) × v × dv/ds

F × ds = (W/g ) × v × dv ------------------(2) Let the initial velocity be u and the final velocity after it moves through a distance ‘ s’ be v

Work-Energy relation for translation 4

Integrating both sides, we get s v F ds = (W/g) v dv 0 u

V

F × s = ( W/g ) v2/2 ] u

Therefore work done by a system of forces acting on a body while causing a displacement is equal to the change in kinetic energy of the body during the displacement.

∑ F × s = (W/2g) [v2-u2]

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Work done by a spring

s

F=k(s)

The force required to cause unit deformation of the spring is called the spring modulus denoted by the symbol k. The force required to deform a spring is given by F= ks. Work done by the force on a spring is the product of the average force and the deformation s.

W= -ks2/2. [ The negative sign indicates that whenever spring is deformed the force of spring is in the opposite direction of deformation.]

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IMPULSE – MOMENTUM

Momentum:-Quantity of motion possessed by a body is called momentum. It is the product of mass and velocity. It is a vector quantity.

Unit:- N s.

Impulse of a Force:- It is defined as the product of force and the time over which it acts. It is a vector quantity.

Unit:- N s.

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Impulse-momentum relationship

Force = Rate of change of momentum F =( mv – mu) /t = Force causing impulse F×t = mv – mu Impulse = final momentum – Initial momentum The component of the resultant linear impulse along any direction is equal to change in the component of momentum in that direction.

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Law of Conservation of momentum

The law of conservation of momentum may be stated as , “momentum is conserved in a system in which resultant force is zero”. In other words, in a system if the resultant is zero, Initial momentum is equal to Final momentum

m1u1+m2u2 = m1v1+m2v2

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ASSIGNMENT1.A small block starts from rest at point a and slides down the inclined

plane. At what distance along the horizontal will it travel before coming to rest . Take µk=0.3 [Ans :s=6m ]

A

3

4

5m

B C

s

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2.The system starts from rest in the position shown . How much further will block ‘A’ move up the incline after block B hits the ground . assume the pulley to be frictionless and massless and µ is 0.2 .WA=1000N, WB=2000N. [ Answer s =1.27m]

3m

A

B34

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3.Two bodies a and b weighing 2000N and 5200N are connected as shown in the figure . find the further distance moved by block a

after the block b hits block c .[ Answer s=1.34m]

512

A

B 3m

4.A 1500Kg automobile travels at a uniform rate of 50kmph to 75kmph . During the entire motion, the automobile is traveling on a level horizontal road and rolling resistance is 2 % of weight of automobile . Find (i) maximum power developed (ii) power required to maintain a constant speed of 75kmph.[ ANSWER: power developed = 6.131KN]

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5.A spring is used to stop 60kg pack age which is sliding on a horizontal surface . the spring has a constant k which is 20kN/m and is held by cable such that it is initially compressed at 120mm. knowing that the package has a velocity of 2.5m/s in position shown and maximum additional displacement of spring is 40mm . Determine the co-eff of kinetic friction between package and surface. (Answer µk=0.2)

600m

60kg

2.5 m/s

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6. The system shown in figure has a rightward velocity of 4m/s, just before force P is applied. Determine the value of P that will give a leftward velocity of 6m/s in a time interval of 20sec. Take µ = 0.2 & assume ideal pulley. [Answer P=645.41N]

1000N

400N

P

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7. A locomotive of weight 500kN pulls a train of weight of 2500kN. The tractive resistance, due to friction is 10N/kN. The train can go with a maximum speed of 27kmph on a grade of 1in100. Determine (a) Power of the locomotive. (b) Maximum speed it can attain on a straight level track with the tractive resistance remaining same. [Answer (a) Power= 450kN (b) v=15m/s]

8.A wagon weighing 400kN starts from rest, runs 30m down a 1% grade & strikes a post. If the rolling resistance of the track is 5N/kN, find the velocity of the wagon when it strikes the post.

If the impact is to be cushioned by means of one bumper string, which compresses 1mm per 20 kN weight, determine how much the bumper spring will be compressed. [Answer v=1.4m/s, x=63.21mm]

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9. A car weighing 11,100 N & running at 10m/s holds 3 men each weighing 700N. The men jumps off from the back end gaining a relative velocity of 5m/s with the car if the three men jumps off (i) in succession, (ii) all together. [Answer (i) v=10.85m/s. (ii) v=10.8m/s.]

10. A gun weighing 300kN fires a 5kN projectile with a velocity of 300m/s. With what velocity will the gun recoil? If the recoil velocity is over come by an average force of 600N, how far will the gun travel & how long will it travel. [Answer v=-5m/s s=0.637m/s t=0.255s ]

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work-energy-power (i) work :- work done by a force is the product of the force and the distance...

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force of spring

product of force

displacement of force

force velocity unit

average force

force distance time

u v f s

s v f ds