Work, Energy, and Power Formulas

Question 1

work =?

Accepted Answer

W=F.S=FScos(theta)

Question 2

1erg , 1eV,1 cal,1kWh how many joule

Accepted Answer

1erg=10^-7J 1eV=1.6*10^-19J 1calorie=4.2J 1kWh=3.6*10^6J

Question 3

work done using work energy theorem

Accepted Answer

wext=∆K.E=(1/2)m((v2)^2-(v1)^2)

Question 4

work done in ping the chain

Accepted Answer

MgL/2(n^2)

Question 5

work done by conservative force formula

Accepted Answer

W=-∆U

Question 6

elastic potential energy

Accepted Answer

U=(1/2)Kx^2=1/2 (Fx)= 1/2 (F^2/k)

Question 7

Force and potential relationship

Accepted Answer

F=-dU/dx

Question 8

stable equilibrium

Accepted Answer

(d^2U/dx^2)>0

Question 9

unstable equilibrium

Accepted Answer

(d^2U/dx^2)<0

Question 10

neutral equilibrium

Accepted Answer

(d^2U/dx^2)=0

Question 11

kinetic energy

Accepted Answer

K.E=(1/2)mv^2=p^2/2m=(1/2)pv

Question 12

p and K.E relation

Accepted Answer

p=√(2mK.E)

Question 13

percentage change in kinetic energy

Accepted Answer

∆K.E/K.E *100% =(Pf^2-Pi^2/Pi^2)*100%

Question 14

percentage change in momentum

Accepted Answer

(∆p/p)*100%=(√K.Ef -√K.Ei/√K.Ei)*100%

Question 15

conservation of mechanical energy

Accepted Answer

Uf+Kf=Ui+Ki

Question 16

conservation of energy for spring mass system

Accepted Answer

k(x(max))^2 =kx^2+mv^2 , vmax=x(max)√(k/m)

Question 17

vertical circular motion Tension

Accepted Answer

T=mv^2/R+mgcos(theta) v is the velocity at that point

Question 18

vertical circular motion Tension with Va the point at the lowest point(special case theta =0°,90°,180°

Accepted Answer

T=m(Va)^2/R+3mgcos(theta)-2mg T(0°)=mva^2/R+mg T(90°)=mva^2/R-2mg T(180°)=mva^2/R-5mg

Question 19

critical velocity (at highest point)

Accepted Answer

vc>or = √(Rg)

Question 20

to complete and full circle from lowest point velocity?

Accepted Answer

V=√(5gR)

Question 21

velocity at lowest point to reach theta=90°

Accepted Answer

v=√(2gR)

Question 22

Va=√(5gR) T at 0° ,90°,180°

Accepted Answer

T(0°)=6mg T(90°)=3mg T(180°)=0

Question 23

power(average, instantaneous,)

Accepted Answer

average=work/time instantaneous=Fvcos(theta) =F.v 1hp=746watt

Question 24

constant power relationship with (x,v,a,t)

Accepted Answer

v=√(2Pt/m) x proportional to t^3/2 v proportional to t^1/2 accleration proportional to t^-1/2

Question 25

conservation of momentum

Accepted Answer

m1u1+m2u2=m1v1+m2v2

Question 26

coefficient of restitution (e)

Accepted Answer

e=|v2-v1|/|u1-u2|

Question 27

loss of kinetic energy with (e)

Accepted Answer

∆K=m1m2/2(m1+m2) *(u1-u2)^2*(1-e^2)

Question 28

perfectly elastic collision in 1D (v1,v2)

Accepted Answer

v1=(m1-m2)u1+2m2u2/(m1+m2) v2=(m2-m1)u2+2m1u1/(m1+m2)

Question 29

for velocity to be interchanged what is the condition of their mass

Accepted Answer

m1=m2

Question 30

if lighter mass colides with heavy target

Accepted Answer

v1=-u1 ,v2=0

Question 31

If heavier r mass colides with lighter target

Accepted Answer

v1=u1 v2=2u1

Question 32

(∆K/K) retained=?

Accepted Answer

(∆K/K) retained= (m1-m2/m1+m2)^2

Question 33

(∆K/K) transferred/lost=?

Accepted Answer

(∆K/K) transferred/lost=4m1m2/(m1+m2)^2

Question 34

(∆K/K) transferred/lost=? if m1=nm2

Accepted Answer

4n/(n+1)^2

Question 35

final velocity of perfectly inelastic collision

Accepted Answer

v=m1v1+m2v2/(m1+m2)

Question 36

elastic collision in 2D conservation of momentum

Accepted Answer

x direction: m1u1=m1v1cos(theta1)+m2v2cos(theta2) y direction : 0=m1v1sin(theta1)-m2v2sin(theta2) here theta1+theta2=90°

Question 37

elastic collision in 2D conservation of momentum (kinetic energy

Accepted Answer

(1/2)m1u1^2=(1/2)m1v1^2+1/2m2v2^2

Question 38

rebound of ball after collision with ground (height , velocity,timetaken, total distance travelled by ball

Accepted Answer

vn= e^n v° , hn=e^2n h° ,d=h°[1+e^2/1-e^2], T=(1+e/1-e)*√(2h°)

Work, Energy, and Power Formulas

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