Solved Examples on Work, Power & Energy

Question 1:-A 1380-kg block of granite is dragged up an incline at a constant speed of 1.34 m/s by a steam winch (below figure). The coefficient of kinetic friction between the block and the incline is 0.41. How much power must be supplied by the winch?

Concept:Weight of an object

Wis equal to the mass of the object times free fall accelerationg.So,

W=mgThe force of friction

fis defined as,

f=µ_{k}NHere

µ_{k}is the coefficient of kinetic friction andNis the normal force.Power

Pof an object is equal to the tension acting on the objectTtimes velocity of the objectv.P=Tv

Solution:The below figure shows a 1380-kg block of granite is dragged up an incline at a constant speed of 1.34 m/s by a steam winch.

Figure:

From the figure,

tan

θ=p/bOr,

θ= tan^{-1}(p/b)To obtain angle

θ, substitute 28.2 m forpand 39.4 m forbin the equationθ= tan^{-1}(p/b),

θ= tan^{-1}(p/b)= tan

^{-1}(p/b)= tan

^{-1}(28.2 m/39.4 m)= tan

^{-1}(28.2 /39.4 )=35.5°

The parallel component of the weight of the block will be,

And the perpendicular component will be,

The force of friction on the block is,

f=µ_{k}NTo obtain the force of friction

fon the block, substitutemgcosθforNin the equationf=µ_{k}N,

f=µ_{k}N=

µ_{k }mgcosθSo the tension

Ton the rope in order to move the block up the ramp will be,To obtain the tension

Ton the rope, substitutemgsinθfor andµ_{k }mgcosθforNin the equation ,So the power

Pthat must be supplied by the winch will be,

P=Tv

=mgv(sinθ+µ_{k}cosθ)To obtain the power

Pthat must be supplied by the winch, substitute 1380 kg for m, 9.81 m/s^{2}for g, 1.34 m/s forv, 0.41 for µ_{k}and 35.5° for θ in the equationP=mgv(sinθ+µ_{k}cosθ),

P=mgv(sinθ+µ_{k}cosθ)= (1380 kg) (9.81 m/s

^{2}) (1.34 m/s) [sin (35.5º) +(0.41) cos (35.5º)]= (1.66×10

^{4}kg.m^{2}/s^{3}) [1 W/(1 kg.m^{2}/s^{3})]= 1.66×10

^{4}WFrom the above observation we conclude that, the power

Pthat must be supplied by the winch would be 1.66×10^{4}W.

Question 2:-The hydrogen-filled Hindenburg (see below figure) could cruise at 77 knots with the engines 4800 hp. Calculate the air drag force in newton’s on the airship at this speed.

:

ConceptPower delivered (

P) by the body is equal to the force acting on the body (F) times speed of the body (v).So,

P=Fv…… (1)From equation (1), the force (

F) will be,

F=P/v…… (2)

Solution:To obtain air drag force in newtons, substitute 77 knots for

vand 4800 hp forPin the equationF=P/v,

F=P/v= 4800 hp/77 knots

= (4800 hp/77 knots) (745.7 W/1 hp) (1 knot/1.688 ft/s) (1 ft/s/0.3048 m/s)

= 9.0×10

^{4}W/m/s= (9.0×10

^{4}W/m/s) [1 N/(1 W/m/s)]=9.0×10

^{4}N …… (3)From the above observation we conclude that, air drag force in newtons on the airship at this speed will be 9.0×10

^{4}N.

Question 3:-A jet airplane is traveling 184 m/s. In each second the engine takes in 68.2 m^{3}of air having a mass of 70.2 kg. The air is used to burn 2.92 kg of fuel on each second. The energy is used to compress the products of combustion and to eject them at the rear of the engine at 497 m/s relative to the plane. Find(a)the thrust of the jet engine and(b)the delivered power (horse power) .

Concept:Force or thrust

Fis equal to the rate of change of momentum (Δp/ Δt).

F= Δp/ ΔtPower

Pis defined as,

P=FvHere

Fis the force andvis the velocity.

Solution:

(a)First we have to find out the change in momentum Δpof the jet engine.The change in momentum Δ

pwill be,Δ

p= (70.2 kg) (497 m/s-184 m/s) +(2.92 kg) (497 m/s= 2.34×10

^{4}kg.m/sTo obtain the thrust

Fof the jet engine, substitute 2.34×10^{4}kg.m/s for Δpand 1 s for Δtin the equationF= Δp/ Δt,

F= Δp/ Δt= (2.34×10

^{4}kg.m/s)/(1 s)= (2.34×10

^{4}kg.m/s^{2}) (1 N/1 kg.m/s^{2})=2.34×10

^{4}NFrom the above observation we conclude that, the thrust

Fof the jet engine would be 2.34×10^{4}N.

(b)To obtain the delivered powerPof the jet engine, substitute 2.34×10^{4}N for thrustFand 184 m/s for the velocityvof the jet airplane in the equationP=Fv, we get,

P=Fv= (2.34×10

^{4}N) (184 m/s)= (4.31×10

^{6}N.m/s) (1 W/1 N.m/s)= 4.31×10

^{6}W= (4.31×10

^{6}W) (0.001341 hp/1 W)= 5780 hp

From the above observation we conclude that, the delivered power

Pof the jet engine would be 5780 hp.

Question 4:-The string in below figure has a lengthL= 120 cm, and the distancedto the fixed peg is 75.0 cm. when the ball is released from rest in the position shown, it will follow the arc shown in the figure. How fast will it be going(a)when it reaches the lowest point in its swing and(b)when it reaches its highest point, after the string catches on the peg?

Concept:Kinetic energy

Kof an object is defined as,

K= ½mv^{2}Here

mis the mass of the object andvis the speed of the object.Potential energy

Uis defined as,

U=mgyHere,

mis the mass of the object,gis the free fall acceleration andyis the height.In accordance to conservation of energy, the total change in kinetic energy Δ

K_{total}of all the particles that make up the system is equal in magnitude, but opposite in sign, to the total change in the potential energy -ΔU_{total}of the system.So, Δ

K_{total}= -ΔU_{total}

K-K_{0}= -(U-U_{0})

Solution:Assume that,

U_{0}=K_{0}= 0Substitute 0 for

U_{0}and 0 forK_{0}in the equationK-K_{0}= -(U-U_{0}), we get,

K-0 = -(U-0)So,

K= -USubstitute ½

mv^{2}for kinetic energyKandmgyfor potential energyUin the equation

K= -U, we get,

K= -U½

mv^{2}= -mgy

v^{2}/2 = -gyOr,

v= √2g(-y)

(a)To obtain the speedvof the ball, when it reaches the lowest point in its swing, substitute 9.81 m/s^{2}for free fall accelerationgand -120 cm fory(negative sign for negative axis) in the equationv= √2g(-y), we get,

V= √2g(-y)=√2(9.81 m/s

^{2})(-(-120 cm))= √2(9.81 m/s

^{2})(120 cm)=√2(9.81 m/s

^{2})(120 cm)(10^{-2}m/1cm)=4.85 m/s

From the above observation we conclude that, the speed

vof the ball, when it reaches the lowest point in its swing would be 4.85 m/s.

(b)The value ofy, when the ball reaches its highest point would be,

y= -[120 cm-(120 cm-75.0 cm)-(120 cm-75.0 cm)]= -[120 cm- 45.0 cm- 45.0 cm]

= (- 30 cm) (10

^{-2}m/1 cm)= -0.30 m

To obtain the speed

vof the ball, when the ball reaches its highest point, substitute 9.81 m/s^{2}for free fall accelerationgand -0.30 m fory(negative sign for negative axis) in the equationv= √2g(-y), we get,

v= √2g(-y)=√2(9.81 m/s

^{2})(-(-0.30 m))= √2(9.81 m/s

^{2})(0.30 m)=2.43 m/s

From the above observation we conclude that, the speed

vof the ball, when the ball reaches its highest point, after the string catches on the peg would be 2.43 m/s.