Physics-
The relation between the displacement 
of an object produced by the application of the variable force 
is represented by a graph shown in the figure. If the object undergoes a displacement from 
to 
the work done will be approximately equal to
Physics-General
Answer:The correct answer is: Work done = Area under curve and displacement axis
= Area of trapezium
As the area actually is not trapezium so work done will be more than 
approximately
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physics-
In the given curved road, if particle is released from 
then
If the surface is smooth then the kinetic energy at 
never be zero
If the surface is rough, the kinetic energy at be zero. Because, work done by force of friction is negative. If work done by friction is equal to 
then, net work done on body will be zero. Hence, net change in kinetic energy is zero. Hence, (b) is correct If the surface is rough, the kinetic energy at must be lesser than . If surface is smooth, the kinetic energy at is equal to The reason is same as in (a) and (b)
never be zeroIf the surface is rough, the kinetic energy at
be zero. Because, work done by force of friction is negative. If work done by friction is equal to then, net work done on body will be zero. Hence, net change in kinetic energy is zero. Hence, (b) is correct If the surface is rough, the kinetic energy at must be lesser than . If surface is smooth, the kinetic energy at is equal to The reason is same as in (a) and (b)In the given curved road, if particle is released from then
physics-General
If the surface is smooth then the kinetic energy at never be zero
If the surface is rough, the kinetic energy at be zero. Because, work done by force of friction is negative. If work done by friction is equal to then, net work done on body will be zero. Hence, net change in kinetic energy is zero. Hence, (b) is correct If the surface is rough, the kinetic energy at must be lesser than . If surface is smooth, the kinetic energy at is equal to The reason is same as in (a) and (b)
never be zeroIf the surface is rough, the kinetic energy at
be zero. Because, work done by force of friction is negative. If work done by friction is equal to then, net work done on body will be zero. Hence, net change in kinetic energy is zero. Hence, (b) is correct If the surface is rough, the kinetic energy at must be lesser than . If surface is smooth, the kinetic energy at is equal to The reason is same as in (a) and (b)physics-
Three objects 
and 
are kept in a straight line on a frictionless horizontal surface. These have masses 
and 
respectively. The object moves towards with a speed 
and makes an elastic collision with it. Thereafter, makes completely inelastic collision with . All motions occur on the same straight line. Find the final speed (in 
) of the object
After elastic collision strikes to with velocity of 
. Now collision between and is perfectly inelastic
By the law of conservation of momentum
Three objects and are kept in a straight line on a frictionless horizontal surface. These have masses and respectively. The object moves towards with a speed and makes an elastic collision with it. Thereafter, makes completely inelastic collision with . All motions occur on the same straight line. Find the final speed (in ) of the object
physics-General
After elastic collision strikes to with velocity of . Now collision between and is perfectly inelasticBy the law of conservation of momentum
physics-
A body of mass 
slides down a curved track which is quadrant of a circle of radius 
. All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is
By conservation of energy, 
A body of mass slides down a curved track which is quadrant of a circle of radius . All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is
physics-General
By conservation of energy,
physics-
The relationship between the force F and position 
of a body is as shown in figure. The work done in displacing the body from 
to 
m will be
Work done=area enclosed by 
graph
=area of ABNM + area of CDEN - area of EFGH + area of HIJ
=
graph=area of ABNM + area of CDEN - area of EFGH + area of HIJ
=
The relationship between the force F and position of a body is as shown in figure. The work done in displacing the body from to m will be
physics-General
Work done=area enclosed by graph
=area of ABNM + area of CDEN - area of EFGH + area of HIJ
=
graph=area of ABNM + area of CDEN - area of EFGH + area of HIJ
=
physics-
A block of mass 
kg sliding on a smooth horizontal surface with a velocity 
meets the spring of spring constant 
fixed at one end as shown in figure. The maximum compression of the spring and velocity of block as is returns to the original position respectively are
When block strikes the spring, the kinetic energy of block converts into potential energy of spring ie,
Or
When block returns to the original position, again potential energy converts into kinetic energy of the blocks, so velocity of the block is same as before but its sign changes as it goes to mean position.
Or
When block returns to the original position, again potential energy converts into kinetic energy of the blocks, so velocity of the block is same as before but its sign changes as it goes to mean position.
A block of mass kg sliding on a smooth horizontal surface with a velocity meets the spring of spring constant fixed at one end as shown in figure. The maximum compression of the spring and velocity of block as is returns to the original position respectively are
physics-General
When block strikes the spring, the kinetic energy of block converts into potential energy of spring ie,
Or
When block returns to the original position, again potential energy converts into kinetic energy of the blocks, so velocity of the block is same as before but its sign changes as it goes to mean position.
Or
When block returns to the original position, again potential energy converts into kinetic energy of the blocks, so velocity of the block is same as before but its sign changes as it goes to mean position.
physics-
A 10 kg brick moves along an -axis. Its acceleration as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from 
to 
m?
According to the graph the acceleration 
varies linearly with the coordinate . We may write 
, where 
is the slope of the graph.
From the graph
The force on the brick is in the positive-direction and according to Newton’s second law, its magnitude is given by
If
is the final coordinate, the work done by the force is
varies linearly with the coordinate . We may write , where is the slope of the graph.From the graph
The force on the brick is in the positive
-direction and according to Newton’s second law, its magnitude is given byIf
is the final coordinate, the work done by the force isA 10 kg brick moves along an -axis. Its acceleration as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from to m?
physics-General
According to the graph the acceleration varies linearly with the coordinate . We may write , where is the slope of the graph.
From the graph
The force on the brick is in the positive-direction and according to Newton’s second law, its magnitude is given by
If is the final coordinate, the work done by the force is
varies linearly with the coordinate . We may write , where is the slope of the graph.From the graph
The force on the brick is in the positive
-direction and according to Newton’s second law, its magnitude is given byIf
is the final coordinate, the work done by the force isphysics-
An object of mass is tied to a string of length 
and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angel 
with the vertical. Work done by the force is
or 
(Since, change in kinetic energy is zero)
Here,
work done by tension = 0
work done by fore of gravity
An object of mass is tied to a string of length and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angel with the vertical. Work done by the force is
physics-General
or (Since, change in kinetic energy is zero)
Here,
work done by tension = 0 work done by fore of gravityphysics-
A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 
and 
. Taking 
, find the work done (in joules) by the string on the block of mass during the first second after the system is released from rest
In the given condition tension in the string
And acceleration of each block
Let ‘S’ is the distance covered by block of mass in first sec
Work done by the string 
And acceleration of each block
Let ‘S’ is the distance covered by block of mass
in first sec Work done by the string A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses and . Taking , find the work done (in joules) by the string on the block of mass during the first second after the system is released from rest
physics-General
In the given condition tension in the string
And acceleration of each block
Let ‘S’ is the distance covered by block of mass in first sec
Work done by the string
And acceleration of each block
Let ‘S’ is the distance covered by block of mass
in first sec Work done by the string maths-
Range of function f(x) = 
, 
is given by -
]
(C)
(C)
Range of function f(x) = , is given by -
maths-General
]
(C)
(C)
physics-
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 
Work = Area under 
graph
graphForce 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
physics-General
Work = Area under graph
graphphysics-
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
, up to distance , constant (negative) and becomes zero suddenlyThe 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
, up to distance , constant (negative) and becomes zero suddenlyphysics-
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
ABCD +Area CEFDThe 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
ABCD +Area CEFDphysics-
Figure shows the -graph. Where is the force applied and is the distance covered
By the body along a straight line path. Given that is in 
and 
in 
, what is the work done?
Work done =area under curve and displacement axis
Figure shows the -graph. Where is the force applied and is the distance covered
By the body along a straight line path. Given that is in and in , what is the work done?
physics-General
Work done =area under curve and displacement axis
physics-
A vertical spring with force constant 
is fixed on a table. A ball of mass at a height 
above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance . The net work done in the process is
Gravitational potential energy of ball gets converted into elastic potential energy of the spring 
Net work done
Net work done
A vertical spring with force constant is fixed on a table. A ball of mass at a height above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance . The net work done in the process is
physics-General
Gravitational potential energy of ball gets converted into elastic potential energy of the spring
Net work done
Net work done
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
, 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
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
, 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