- Work done Formula
- Work done by variable force
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- Kinetic energy
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- Work Energy theorem
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- Potential energy
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- Potential energy of the spring
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- Conservative Forces
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- Mechanical Energy and Conservation
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- Power definition
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- Law of Conservation of energy
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- Work Energy and Power Solved Examples

t

a. $t_1 > t_2$

b. $t_1 < t_2$

c. $t_1 = t_2$

d. Depends on the mass of the body

Let H be the height

then

First Half

$\frac {H}{2}=\frac {1}{2}gt_1^2$ ----(1)

Also $v=gt_1$

Second Half

$\frac {H}{2}=vt_2+\frac {1}{2}gt_2^2$

$\frac {H}{2}=gt_1t_2+\frac {1}{2}gt_2^2$

or

$\frac {1}{2}gt_2^2 =\frac {H}{2}-gt_1t_2$ ---(2)

From 1 and 2

$\frac {1}{2}gt_2^2=\frac {1}{2}gt_1^2 -gt_1t_2$

$t_2^2+2t_1t_2 -t_1^2=0$

or $t_2=\frac {[-2t_1+ \sqrt {(4t_1^2 + 4t_1^2)}]}{2}$

or $t_2=\frac {[-2t_1+2t_1 \sqrt {2}]}{2}$

$t_2=.44t_1$

so, $t_1 > t_2$

Hence a is correct

a. Work done between two points is independent of the path

b. Work done in a closed loop is zero

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

d. None of the these

Work done between two points is independent of the path

Work done in a closed loop is zero

And $-W=U_f-U_i$

So for positive work potential energy decreases

So a and b are correct

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

b.The period depends on the amplitude of the oscillation

c.the period is independent of mass m

d. the period is independent of length 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. $\sqrt {gl}$

b. $\sqrt {2gl(1-cos \theta)}$

c. $\sqrt {2glcos \theta}$

d.$\sqrt{2gl(1-sin \theta)}$

$h=l-lcos \theta$

or $h=l(1-cos \theta)$

So potential energy at this point is given by

$=mgh=mgl(1-cos \theta)$

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

if v be the velocity of the bob then Kinetic Energy=$\frac {1}{2}mv^2$

Now $\frac {1}{2}mv^2=mgl(1-cos \theta)$

or $v=\sqrt {2gl(1-cos \theta)}$

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 $-\mu Ns$ where μ is coefficient of friction and N is normal reaction.

Let h be the height of the incline plane then the gravitational 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 kinetic energy at the initial point= $\frac {1}{2}mv^2$

Now, applying work energy theorem

$K_f-K_i=W_{grav} + W_{fric}$

Now since K

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

Therefore

$-\frac {1}{2}mv^2=-mgh-\mu Ns $

$\frac {1}{2}mv^2=mgh + \mu Ns $

Now s=3

$N=mgcos 20$

$h=s sin 20$

Substituting all the values

$\frac {1}{2}mv^2=42.2J$

a.if Work done 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 work done by the conservative force in closed loop is zero

d. None of the above

The work done by a conservative force is equal to the negative of the potential energy.When the work done 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 work done by the conservative system is zero in closed loop

Hence a,b,c are correct

$U=20x^2+35z^3$

Find the vector force on it

a. -40x

b. 40x

c.-10x

d. 40x

$U=20x^2+35z^3$

$\mathbf{F}=-\frac{\partial U}{\partial x}\mathbf{i} -\frac{\partial U}{\partial y}\mathbf{j} -\frac{\partial U}{\partial z}\mathbf{k}$

or

a. Frictional force

b. Gravitational force

c. Electrical force

d. Viscous force

P. Work done by the force in closed loop is zero

Q. Work done 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, Work done by the force in closed loop is zero

And for non conservative force,Work done 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. Velocity 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. Work done

b. Kinetic Energy

c. Potential Energy

d. Force

Work done Unit is joule

KE and PE also unit is Joule

Force Unit is Newton

So Solution (d)

P)Kinetic Energy

Q)Potential Energy

R)Momentum

S)Mechanical Energy

A)Mass

B)Velocity

C)Position of the object

D)Volume

Kinetic Energy=$\frac {1}{2}mv^2$

Potential Energy=mgh

Momentum=mv

Mechanical Energy= Kinetic Energy + Potential Energy

P -> A,B

Q-> A,C

R-> A,B

S-> A,B.C

a. Potential Energy of the truck does not change b. Final Kinetic Energy of the truck is $2Mv^2$

c. Work done by the acceleration peddle is $1.5Mv^2$

d. Potential Energy of the truck becomes four times of the initial Potential energy

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

Initial Kinetic Energy =$\frac {1}{2}Mv^2$

Final Kinetic Energy=$2Mv^2$

Work done by the acceleration peddle= Final KE - Initial KE=$1.5Mv^2$

Class 11 Maths Class 11 Physics Class 11 Chemistry

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