Physics-
General
Easy
Question
Force
on a particle moving in a straight line varies with distance
as shown in the figure. The work done on the particle during its displacement of 

- 13 J
- 18 J
- 21 J
- 26 J
The correct answer is: 13 J
Work = Area under
graph

Related Questions to study
physics-
The potential energy of a system is represented in the first figure. The force acting on the system will be represented by

As slope of problem graph is positive and constant upto certain distance and then it becomes zero
So from
, up to distance
,
constant (negative) and becomes zero suddenly
So from
The potential energy of a system is represented in the first figure. The force acting on the system will be represented by

physics-General
As slope of problem graph is positive and constant upto certain distance and then it becomes zero
So from
, up to distance
,
constant (negative) and becomes zero suddenly
So from
physics-
The work done by force acting on a body is as shown in the graph. The total work done in covering an initial distance of 20 m is

Work done 
ABCD +Area CEFD




The work done by force acting on a body is as shown in the graph. The total work done in covering an initial distance of 20 m is

physics-General
Work done 
ABCD +Area CEFD




physics-
Two rectangular blocks
and
of masses 2kg and 3 kg respectively are connected by spring of spring constant 10.8
and are placed on a frictionless horizontal surface. The block
was given an initial velocity of 0.15
in the direction shown in the figure. The maximum compression of the spring during the motion is

As the block A moves with velocity with velocity 0.15
, it compresses the spring Which pushes B towards right. A goes on compressing the spring till the velocity acquired by B becomes equal to the velocity of A, i.e. 0.15
. Let this velocity be v. Now, spring is in a state of maximum compression. Let x be the maximum compression at this stage.

According to the law of conservation of linear momentum, we get

Or

According to the law of conservation of energy





Or

According to the law of conservation of linear momentum, we get
Or
According to the law of conservation of energy
Or
Two rectangular blocks
and
of masses 2kg and 3 kg respectively are connected by spring of spring constant 10.8
and are placed on a frictionless horizontal surface. The block
was given an initial velocity of 0.15
in the direction shown in the figure. The maximum compression of the spring during the motion is

physics-General
As the block A moves with velocity with velocity 0.15
, it compresses the spring Which pushes B towards right. A goes on compressing the spring till the velocity acquired by B becomes equal to the velocity of A, i.e. 0.15
. Let this velocity be v. Now, spring is in a state of maximum compression. Let x be the maximum compression at this stage.

According to the law of conservation of linear momentum, we get

Or

According to the law of conservation of energy





Or

According to the law of conservation of linear momentum, we get
Or
According to the law of conservation of energy
Or
physics-
Six identical balls are linked in a straight groove made on a horizontal frictionless surface as shown. Two similar balls each moving with a velocity
collide elastically with the row of 6 balls from left. What will happen

Momentum and kinetic energy is conserved only in this case
Six identical balls are linked in a straight groove made on a horizontal frictionless surface as shown. Two similar balls each moving with a velocity
collide elastically with the row of 6 balls from left. What will happen

physics-General
Momentum and kinetic energy is conserved only in this case
physics-
A force
acting on an object varies with distance
as shown here. The force is in
and
in
. The work done by the force in moving the object from
to
is

Work done = Area enclosed by
graph

A force
acting on an object varies with distance
as shown here. The force is in
and
in
. The work done by the force in moving the object from
to
is

physics-General
Work done = Area enclosed by
graph

physics-
What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of
(Take
)

What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of
(Take
)

physics-General
physics-
The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant k and compresses it by length L. The maximum momentum of the block after collides is
Momentum would be maximum when KE would be maximum and this is the case when total elastic PE is converted KE.
According to conservation of energy

Or


According to conservation of energy
Or
The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant k and compresses it by length L. The maximum momentum of the block after collides is
physics-General
Momentum would be maximum when KE would be maximum and this is the case when total elastic PE is converted KE.
According to conservation of energy

Or


According to conservation of energy
Or
physics
The area covered by the curve of V-t graph and time axis is equal to magnitude of
The area covered by the curve of V-t graph and time axis is equal to magnitude of
physicsGeneral
physics-
In a children’s park, there is a slide which has a total length of 10 m and a height of 8.0 m. A vertical ladder is provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three-tenth of his weight. The work done by the slide on the boy as he comes down is

or
In a children’s park, there is a slide which has a total length of 10 m and a height of 8.0 m. A vertical ladder is provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three-tenth of his weight. The work done by the slide on the boy as he comes down is

physics-General
or
physics-
A mass ‘m’ moves with a velocity 'v’ and collides in elastically with another identical mass. After collision the Ist mass moves with velocity
in a direction perpendicular to the initial direction of motion. Find the speed of the 2nd mass after collision

Let mass
moves with velocity
and collides inelastically with mass
which is at rest

According to problem mass
moves in a perpendicular direction and let the mass
moves at angle
with the horizontal with velocity 
Initial horizontal momentum of system
(before collision)
….(i)
Final horizontal momentum of system
(after collision)
….(ii)
From the conservation of horizontal linear momentum
…(iii)
Initial vertical momentum of system (before collision) is zero
Final vertical momentum of system
From the conservation of vertical linear momentum
…(iv)
By solving (iii) and (iv)



According to problem mass
Initial horizontal momentum of system
(before collision)
Final horizontal momentum of system
(after collision)
From the conservation of horizontal linear momentum
Initial vertical momentum of system (before collision) is zero
Final vertical momentum of system
From the conservation of vertical linear momentum
By solving (iii) and (iv)
A mass ‘m’ moves with a velocity 'v’ and collides in elastically with another identical mass. After collision the Ist mass moves with velocity
in a direction perpendicular to the initial direction of motion. Find the speed of the 2nd mass after collision

physics-General
Let mass
moves with velocity
and collides inelastically with mass
which is at rest

According to problem mass
moves in a perpendicular direction and let the mass
moves at angle
with the horizontal with velocity 
Initial horizontal momentum of system
(before collision)
….(i)
Final horizontal momentum of system
(after collision)
….(ii)
From the conservation of horizontal linear momentum
…(iii)
Initial vertical momentum of system (before collision) is zero
Final vertical momentum of system
From the conservation of vertical linear momentum
…(iv)
By solving (iii) and (iv)



According to problem mass
Initial horizontal momentum of system
(before collision)
Final horizontal momentum of system
(after collision)
From the conservation of horizontal linear momentum
Initial vertical momentum of system (before collision) is zero
Final vertical momentum of system
From the conservation of vertical linear momentum
By solving (iii) and (iv)
physics-
A toy car of mass 5 kg moves up a ramp under the influence of force
plotted against displacement
. The maximum height attained is given by

Work done = Gain in potential energy
Area under curve

Area under curve
A toy car of mass 5 kg moves up a ramp under the influence of force
plotted against displacement
. The maximum height attained is given by

physics-General
Work done = Gain in potential energy
Area under curve

Area under curve
physics
In uniformly accelerated motion the slope of velocity - time graph gives ....
In uniformly accelerated motion the slope of velocity - time graph gives ....
physicsGeneral
physics
The graph of displacement (x)
time (t) for an object is given in the figure. In which part of the graph the acceleration of the particle is positive ?

The graph of displacement (x)
time (t) for an object is given in the figure. In which part of the graph the acceleration of the particle is positive ?

physicsGeneral
physics
In the above figure acceleration (a)
time (t) graph is given. Hence 

In the above figure acceleration (a)
time (t) graph is given. Hence 

physicsGeneral
physics-
A block of mass 2kg is free to move along the x- axis. It is at rest and from t=0 onwards it is subjected to a time-dependent force
in the x- direction. The force
varies with t as shown in the figure. The kinetic energy of the block after 4.5 seconds is

A block of mass 2kg is free to move along the x- axis. It is at rest and from t=0 onwards it is subjected to a time-dependent force
in the x- direction. The force
varies with t as shown in the figure. The kinetic energy of the block after 4.5 seconds is

physics-General