- Introduction
- Work
- Work done by variable force
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- Mechanical Energy
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- Kinetic energy
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- Potential energy
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- Gravitational PE near the surface of the earth
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- Conversion of Gravitational PE to KE
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- PE of the spring
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- Conservation of energy of spring mass system
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- Conservative Forces
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- Power
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- Principle of conservation of energy
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- Solved examples

t

a. t

b. t

c t

d Depends on the mass of the body

Let H be the height

then

First Half

H/2=(1/2)gt

or

(1/2)gt

Also v=gt

Second Half

H/2=vt

or

(H/2)=gt

or

(1/2)gt

From 1 and 2

(1/2)gt

t

or t

or t

t

so t

Hence a is correct

a. Workdone between two points is independent of the path

b. Workdone in a closed loop is zero

c. if the workdone by the conservative is positive,its potential energy increases

d. None of the these

Workdone between two points is independent of the path

Workdone in a closed loop is zero

And -W=U

So for positive work potential energy decreases

So a and b are correct

a.The freqiency is independent of the acceleration due to gravity g

b.The period depends on the amplitude of the ocsillation

c.the period is independent of mass m

d. the period is independent of lenght l

So it is independent of the mass

a.v

b.v

c. v

d. v

Total Energy at height h

=(1/2)mv

Total energy at height h

=(1/2)mv

Since we know that total energy remains constant during a free fall

Total Energy at height h

or (1/2)mv

or v

a.&radic:2gl

b. &radic:2gl(1-cosθ)

c.&radic:2glcosθ

d.&radic:2gl(1-sinθ

h=l-lcosθ

or h=l(1-cosθ)

So potential energy at this point is given by

=mgh=mgl(1-cosθ)

When the bob passes through equilibrium position,this potential energy is converted into kinectic energy

if v be the velocity of the bob the KE=(1/2)mv

Now (1/2)mv

or v=radic:2gl(1-cosθ)

a 40 J

b. 42.2 J

c. 42.6 J

d. 45 J

If the package travels the entire length s of the incline ,the frictional force will perform work -μNs where μ is coefficient of friction and N is normal reaction.

Let h be the height of the incline plane then the gravtational potential energy of the package will increase by mgh.

Now let assume v speed be given to the package so as to reach the top

Then kinectic energy at the intial point=(1/2)mv

Now applying work energy thoerm

K.E

Now since K.E

Also Workdone by the gravitational force=-(change in gravitational potential energy)=-mgh

Therefore

-(1/2)mv

or (1/2)mv

Now s=3

N=mgcosθ

h=ssinθ

Substituting all the values

(1/2)mv

a.if Workdone by the conservative force is positive then Potential energy decreases

b. Rate of change of momentum of many particles system is proportional to net external force on the system

c.The workdone by the conservative force in closed loop is zero

d. None of the above

The workdone by a conservative force is equal to the negative of the potential energy.When the wokdone is positive ,the potential energy decreases.

The rate of change of total momentum of a many -particle system is proportional to the net force external to the system ;the internal forces between particles cannot change the momentum of the system.The workdone by the conservative system is zero in closed loop

Hence a,b,c are correct

U=20x

a. -40xi-105z

b. 40xi-105z

c.-10xi-105z

d 40xi+105z

U=20x

F=-(∂/∂x)i--(∂/∂y)j--(∂/∂z)k

or

F=-40xi-105z

a. Frictional force

b. Gravitational force

c. Electrical force

d Viscous force

P. Workdone by the force in closed loop is zero

Q. Workdone by the force in closed loop is not zero

Frictional force is non conservative force

Gravitational force is conservative force

Electrical force is conservative force

Viscous force is non conservative force

And for conservative force, Workdone by the force in closed loop is zero

And for non conservative force,Workdone by the force in closed loop is not zero

a. A train which speeding Up

b. A train with constant speed

c. A train which speeding down

d A train at rest

a. Velcoity is perpendicular to acceleration at the highest point

b. Horizontal components of velocity remains constant through out the path

c. Range of the projectile is given by Horizontal component of velocity X Time of flight

d. None of the above

a)Pm/m+M

b)P

c)PM/m+M

d)Pm/M-m

(A) 24 N

(B) 74 N

(C) 15 N

(D) 49 N

x=t

Find the acceleration of the particle at displacement equal to zero

a.(-8,-2,10)

b. (-1,-2,10)

c. (8,2,10)

d. (1,2,10)

a) Workdone

b) Kinetic Energy

c) Potential Energy

d) Force

Workdone Unit is joule

KE and PE also unit is Joule

Force Unit is Newton

So Solution (d)

P)KE

Q)Potential Energy

R)Momentum

S)Mechanical Energy

A)Mass

B)Velocity

C)Position of the object

D)Volume

KE=.5mv

PE=mgh

Momentum=mv

Mechanical Energy= KE +PE

P -> A,B

Q-> A,C

R-> A,B

S-> A,B.C

a) PE of the truck does not change b) Final KE of the truck is 2Mv

c) Workdone by the acceleration peddle is 1.5Mv

d)PE of the truck becomes four times of the initial Potential energy

Solution : Since it is running on level road. PE does not change

Initial KE =.5Mv

Final KE=2Mv

Workdone by the acceleration peddle= Final KE – Initial KE=1.5Mv

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Class 11 Maths Class 11 Physics Class 11 Chemistry