General
Easy
Physics-

The relation between the displacement X of an object produced by the application of the variable force F is represented by a graph shown in the figure. If the object undergoes a displacement from X equals 0.5 blank m to X equals 2.5 blank m the work done will be approximately equal to

Physics-General

  1. 1.6 blank J    
  2. 8 blank J    
  3. 16 blank J    
  4. 32 blank J    

    Answer:The correct answer is: 16 blank JWork done = Area under curve and displacement axis
    = Area of trapezium
    equals fraction numerator 1 over denominator 2 end fraction cross times open parentheses s u m blank o f blank t w o blank p a r a l l e l blank l i n e s close parentheses cross times d i s t a n c e blank b e t w e e n blank t h e m
    equals fraction numerator 1 over denominator 2 end fraction open parentheses 10 plus 4 close parentheses cross times open parentheses 2.5 minus 0.5 close parentheses equals fraction numerator 1 over denominator 2 end fraction 14 cross times 2 equals 14 blank J
    As the area actually is not trapezium so work done will be more than 14 blank J i. e. approximately 16 blank J

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    Related Questions to study

    General
    physics-

    In the given curved road, if particle is released from A then

    If the surface is smooth then the kinetic energy at B never be zero
    If the surface is rough, the kinetic energy at B be zero. Because, work done by force of friction is negative. If work done by friction is equal to m g h 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 B must be lesser than m g h. If surface is smooth, the kinetic energy at B is equal to m g h The reason is same as in (a) and (b)

    In the given curved road, if particle is released from A then

    physics-General
    If the surface is smooth then the kinetic energy at B never be zero
    If the surface is rough, the kinetic energy at B be zero. Because, work done by force of friction is negative. If work done by friction is equal to m g h 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 B must be lesser than m g h. If surface is smooth, the kinetic energy at B is equal to m g h The reason is same as in (a) and (b)
    General
    physics-

    Three objects A comma B and C are kept in a straight line on a frictionless horizontal surface. These have masses m comma blank 2 m and m respectively. The object A moves towards B with a speed 9 blank m divided by s and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m divided by s) of the object C


    v subscript 2 end subscript equals fraction numerator 2 m subscript 1 end subscript v subscript 1 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 2 cross times m cross times 9 over denominator m plus 2 m end fraction equals 6 blank m divided by s
    i. e. After elastic collision B strikes to C with velocity of 6 blank m divided by s. Now collision between B and C is perfectly inelastic

    By the law of conservation of momentum
    2 m cross times 6 plus 0 equals 3 m cross times v subscript s y s end subscript
    rightwards double arrow v subscript s y s end subscript equals 4 blank m divided by s

    Three objects A comma B and C are kept in a straight line on a frictionless horizontal surface. These have masses m comma blank 2 m and m respectively. The object A moves towards B with a speed 9 blank m divided by s and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m divided by s) of the object C

    physics-General

    v subscript 2 end subscript equals fraction numerator 2 m subscript 1 end subscript v subscript 1 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 2 cross times m cross times 9 over denominator m plus 2 m end fraction equals 6 blank m divided by s
    i. e. After elastic collision B strikes to C with velocity of 6 blank m divided by s. Now collision between B and C is perfectly inelastic

    By the law of conservation of momentum
    2 m cross times 6 plus 0 equals 3 m cross times v subscript s y s end subscript
    rightwards double arrow v subscript s y s end subscript equals 4 blank m divided by s
    General
    physics-

    A body of mass 2 blank k g slides down a curved track which is quadrant of a circle of radius 1 blank m e t r e. All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is

    By conservation of energy, m g h equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent
    rightwards double arrow v equals square root of 2 g h end root equals square root of 2 cross times 9.8 cross times 1 end root equals square root of 19.6 end root equals 4.43 blank m divided by s

    A body of mass 2 blank k g slides down a curved track which is quadrant of a circle of radius 1 blank m e t r e. 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, m g h equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent
    rightwards double arrow v equals square root of 2 g h end root equals square root of 2 cross times 9.8 cross times 1 end root equals square root of 19.6 end root equals 4.43 blank m divided by s
    General
    physics-

    The relationship between the force F and position x of a body is as shown in figure. The work done in displacing the body from x equals 1 mto x equals 5m will be

    Work done=area enclosed by F minus xgraph
    =area of ABNM + area of CDEN - area of EFGH + area of HIJ

    =1 cross times 10 plus 1 cross times 5 minus 1 cross times 5 plus fraction numerator 1 over denominator 2 end fraction cross times 1 cross times 10
    equals 10 plus 5 minus 5 plus 5 equals 15 blank J

    The relationship between the force F and position x of a body is as shown in figure. The work done in displacing the body from x equals 1 mto x equals 5m will be

    physics-General
    Work done=area enclosed by F minus xgraph
    =area of ABNM + area of CDEN - area of EFGH + area of HIJ

    =1 cross times 10 plus 1 cross times 5 minus 1 cross times 5 plus fraction numerator 1 over denominator 2 end fraction cross times 1 cross times 10
    equals 10 plus 5 minus 5 plus 5 equals 15 blank J
    General
    physics-

    A block of mass m equals 25kg sliding on a smooth horizontal surface with a velocity v equals 3 m s to the power of negative 1 end exponent meets the spring of spring constant k equals 100 N m to the power of negative 1 end exponent 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,
    fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    Or x equals square root of fraction numerator m v to the power of 2 end exponent over denominator k end fraction end root
    equals square root of fraction numerator 25 cross times 3 to the power of 2 end exponent over denominator 100 end fraction end root square root of fraction numerator 9 over denominator 4 end fraction end root equals fraction numerator 3 over denominator 2 end fraction equals 1.5 blank m
    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.
    H e n c e v equals negative 3 m s to the power of negative 1 end exponent

    A block of mass m equals 25kg sliding on a smooth horizontal surface with a velocity v equals 3 m s to the power of negative 1 end exponent meets the spring of spring constant k equals 100 N m to the power of negative 1 end exponent 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,
    fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    Or x equals square root of fraction numerator m v to the power of 2 end exponent over denominator k end fraction end root
    equals square root of fraction numerator 25 cross times 3 to the power of 2 end exponent over denominator 100 end fraction end root square root of fraction numerator 9 over denominator 4 end fraction end root equals fraction numerator 3 over denominator 2 end fraction equals 1.5 blank m
    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.
    H e n c e v equals negative 3 m s to the power of negative 1 end exponent
    General
    physics-

    A 10 kg brick moves along an x-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 x equals 0 to x equals 8.0m?

    According to the graph the acceleration a varies linearly with the coordinate x. We may write a equals alpha x, where alpha is the slope of the graph.
    From the graph
    alpha equals fraction numerator 20 over denominator 8 end fraction m g subscript 0 end subscript equals 2.5 blank s to the power of negative 2 end exponent
    The force on the brick is in the positive x-direction and according to Newton’s second law, its magnitude is given by
    F equals fraction numerator a over denominator m end fraction equals fraction numerator alpha over denominator m end fraction x
    If x subscript f end subscript is the final coordinate, the work done by the force is
    W equals not stretchy integral from 0 to x subscript f end subscript of F blank d x equals fraction numerator a over denominator m end fraction not stretchy integral subscript 0 end subscript superscript x subscript f end subscript end superscript x blank d x
    equals fraction numerator alpha over denominator 2 m end fraction x subscript f end subscript superscript 2 end superscript equals fraction numerator 2.5 over denominator 2 cross times 10 end fraction cross times open parentheses 8 close parentheses to the power of 2 end exponent
    equals 8 blank J

    A 10 kg brick moves along an x-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 x equals 0 to x equals 8.0m?

    physics-General
    According to the graph the acceleration a varies linearly with the coordinate x. We may write a equals alpha x, where alpha is the slope of the graph.
    From the graph
    alpha equals fraction numerator 20 over denominator 8 end fraction m g subscript 0 end subscript equals 2.5 blank s to the power of negative 2 end exponent
    The force on the brick is in the positive x-direction and according to Newton’s second law, its magnitude is given by
    F equals fraction numerator a over denominator m end fraction equals fraction numerator alpha over denominator m end fraction x
    If x subscript f end subscript is the final coordinate, the work done by the force is
    W equals not stretchy integral from 0 to x subscript f end subscript of F blank d x equals fraction numerator a over denominator m end fraction not stretchy integral subscript 0 end subscript superscript x subscript f end subscript end superscript x blank d x
    equals fraction numerator alpha over denominator 2 m end fraction x subscript f end subscript superscript 2 end superscript equals fraction numerator 2.5 over denominator 2 cross times 10 end fraction cross times open parentheses 8 close parentheses to the power of 2 end exponent
    equals 8 blank J
    General
    physics-

    An object of mass m is tied to a string of length L and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angel theta with the vertical. Work done by the force F is

    W equals increment K or W subscript T end subscript plus W subscript g end subscript plus W subscript F end subscript equals 0
    (Since, change in kinetic energy is zero)

    Here, W subscript T end subscript equals work done by tension = 0
    W subscript g end subscript equals work done by fore of gravity
    equals negative m g h
    equals negative m g L left parenthesis 1 minus cos invisible function application theta right parenthesis
    therefore blank W subscript F end subscript equals negative W subscript g end subscript equals m g L left parenthesis 1 minus cos invisible function application theta right parenthesis

    An object of mass m is tied to a string of length L and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angel theta with the vertical. Work done by the force F is

    physics-General
    W equals increment K or W subscript T end subscript plus W subscript g end subscript plus W subscript F end subscript equals 0
    (Since, change in kinetic energy is zero)

    Here, W subscript T end subscript equals work done by tension = 0
    W subscript g end subscript equals work done by fore of gravity
    equals negative m g h
    equals negative m g L left parenthesis 1 minus cos invisible function application theta right parenthesis
    therefore blank W subscript F end subscript equals negative W subscript g end subscript equals m g L left parenthesis 1 minus cos invisible function application theta right parenthesis
    General
    physics-

    A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 blank k gand 0.72 blank k g. Taking g equals 10 blank m divided by s to the power of 2 end exponent, find the work done (in joules) by the string on the block of mass 0.36 blank k g during the first second after the system is released from rest

    In the given condition tension in the string

    T equals fraction numerator 2 m subscript 1 end subscript m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction g equals fraction numerator 2 cross times 0.36 cross times 0.72 over denominator 1.08 end fraction cross times 10
    T equals 4.8 blank N
    And acceleration of each block
    a equals open parentheses fraction numerator m subscript 2 end subscript minus m subscript 1 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses g equals open parentheses fraction numerator 0.72 minus 0.36 over denominator 0.72 plus 0.36 end fraction close parentheses g equals fraction numerator 10 over denominator 3 end fraction blank m divided by s to the power of 2 end exponent
    Let ‘S’ is the distance covered by block of mass 0.36 blank k g in first sec
    S equals u t plus fraction numerator 1 over denominator 2 end fraction blank a t to the power of 2 end exponent rightwards double arrow S equals 0 plus fraction numerator 1 over denominator 2 end fraction open parentheses fraction numerator 10 over denominator 3 end fraction close parentheses cross times 1 to the power of 2 end exponent equals fraction numerator 10 over denominator 6 end fraction blank m e t e r
    therefore Work done by the string W equals T S equals 4.8 cross times fraction numerator 10 over denominator 6 end fraction
    rightwards double arrow W equals 8 blank J o u l e

    A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 blank k gand 0.72 blank k g. Taking g equals 10 blank m divided by s to the power of 2 end exponent, find the work done (in joules) by the string on the block of mass 0.36 blank k g during the first second after the system is released from rest

    physics-General
    In the given condition tension in the string

    T equals fraction numerator 2 m subscript 1 end subscript m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction g equals fraction numerator 2 cross times 0.36 cross times 0.72 over denominator 1.08 end fraction cross times 10
    T equals 4.8 blank N
    And acceleration of each block
    a equals open parentheses fraction numerator m subscript 2 end subscript minus m subscript 1 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses g equals open parentheses fraction numerator 0.72 minus 0.36 over denominator 0.72 plus 0.36 end fraction close parentheses g equals fraction numerator 10 over denominator 3 end fraction blank m divided by s to the power of 2 end exponent
    Let ‘S’ is the distance covered by block of mass 0.36 blank k g in first sec
    S equals u t plus fraction numerator 1 over denominator 2 end fraction blank a t to the power of 2 end exponent rightwards double arrow S equals 0 plus fraction numerator 1 over denominator 2 end fraction open parentheses fraction numerator 10 over denominator 3 end fraction close parentheses cross times 1 to the power of 2 end exponent equals fraction numerator 10 over denominator 6 end fraction blank m e t e r
    therefore Work done by the string W equals T S equals 4.8 cross times fraction numerator 10 over denominator 6 end fraction
    rightwards double arrow W equals 8 blank J o u l e
    General
    maths-

    Range of function f(x) = fraction numerator x to the power of 2 end exponent plus 2 x plus 3 over denominator x end fraction, x element of times R is given by -

    ]
    (C) left parenthesis negative infinity comma 2 minus 2 square root of 3 right square bracket union left square bracket 2 plus 2 square root of 3 comma infinity right parenthesis

    Range of function f(x) = fraction numerator x to the power of 2 end exponent plus 2 x plus 3 over denominator x end fraction, x element of times R is given by -

    maths-General
    ]
    (C) left parenthesis negative infinity comma 2 minus 2 square root of 3 right square bracket union left square bracket 2 plus 2 square root of 3 comma infinity right parenthesis
    General
    physics-

    Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 blank m

    Work = Area under left parenthesis F minus d right parenthesis graph
    equals 8 plus 5 equals 13 blank J

    Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 blank m

    physics-General
    Work = Area under left parenthesis F minus d right parenthesis graph
    equals 8 plus 5 equals 13 blank J
    General
    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 F equals fraction numerator negative d U over denominator d x end fraction, up to distance a,
    F equals constant (negative) and becomes zero suddenly

    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 F equals fraction numerator negative d U over denominator d x end fraction, up to distance a,
    F equals constant (negative) and becomes zero suddenly
    General
    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 W equals A r e a blank A B C E F D A
    equals A r e aABCD +Area CEFD

    equals fraction numerator 1 over denominator 2 end fraction cross times open parentheses 15 plus 10 close parentheses cross times 10 plus fraction numerator 1 over denominator 2 end fraction cross times left parenthesis 10 plus 20 right parenthesis cross times 5
    equals 125 plus 75 equals 200 J

    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 W equals A r e a blank A B C E F D A
    equals A r e aABCD +Area CEFD

    equals fraction numerator 1 over denominator 2 end fraction cross times open parentheses 15 plus 10 close parentheses cross times 10 plus fraction numerator 1 over denominator 2 end fraction cross times left parenthesis 10 plus 20 right parenthesis cross times 5
    equals 125 plus 75 equals 200 J
    General
    physics-

    Figure shows the F-xgraph. Where F is the force applied and x is the distance covered

    By the body along a straight line path. Given that F is in n e w t o n and x blankin m e t r e, what is the work done?

    Work done =area under curve and displacement axis
    equals 1 cross times 10 minus 1 cross times 10 plus 1 cross times 10 equals 10 blank J

    Figure shows the F-xgraph. Where F is the force applied and x is the distance covered

    By the body along a straight line path. Given that F is in n e w t o n and x blankin m e t r e, what is the work done?

    physics-General
    Work done =area under curve and displacement axis
    equals 1 cross times 10 minus 1 cross times 10 plus 1 cross times 10 equals 10 blank J
    General
    physics-

    A vertical spring with force constant K is fixed on a table. A ball of mass mat a height h above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance d. The net work done in the process is

    Gravitational potential energy of ball gets converted into elastic potential energy of the spring m g open parentheses h plus d close parentheses equals fraction numerator 1 over denominator 2 end fraction K d to the power of 2 end exponent
    Net work done equals m g open parentheses h plus d close parentheses minus fraction numerator 1 over denominator 2 end fraction K d to the power of 2 end exponent equals 0

    A vertical spring with force constant K is fixed on a table. A ball of mass mat a height h above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance d. The net work done in the process is

    physics-General
    Gravitational potential energy of ball gets converted into elastic potential energy of the spring m g open parentheses h plus d close parentheses equals fraction numerator 1 over denominator 2 end fraction K d to the power of 2 end exponent
    Net work done equals m g open parentheses h plus d close parentheses minus fraction numerator 1 over denominator 2 end fraction K d to the power of 2 end exponent equals 0
    General
    physics-

    Two rectangular blocks A blankand B blankof masses 2kg and 3 kg respectively are connected by spring of spring constant 10.8 N m to the power of negative 1 end exponentand are placed on a frictionless horizontal surface. The block A blankwas given an initial velocity of 0.15 m s to the power of negative 1 end exponent 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 m s to the power of negative 1 end exponent, 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 m s to the power of negative 1 end exponent. 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
    m subscript A end subscript u equals left parenthesis m subscript A end subscript plus m subscript B end subscript right parenthesis v
    Or v equals fraction numerator m subscript A end subscript u over denominator m subscript A end subscript plus m subscript B end subscript end fraction
    fraction numerator 2 cross times 0.15 over denominator 2 plus 3 end fraction equals 0.06 m s to the power of negative 1 end exponent
    According to the law of conservation of energy
    fraction numerator 1 over denominator 2 end fraction m subscript A end subscript u to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction open parentheses m subscript A end subscript plus m subscript B end subscript close parentheses V to the power of 2 end exponent plus fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    fraction numerator 1 over denominator 2 end fraction m subscript A end subscript u to the power of 2 end exponent minus fraction numerator 1 over denominator 2 end fraction open parentheses m subscript A end subscript plus m subscript B end subscript close parentheses v to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    fraction numerator 1 over denominator 2 end fraction cross times 2 cross times open parentheses 0.15 close parentheses to the power of 2 end exponent minus fraction numerator 1 over denominator 2 end fraction open parentheses 2 plus 3 close parentheses open parentheses 0.06 close parentheses to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    0.0225 minus 0.009 equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    o r blank 0.0135 equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    Or x equals square root of fraction numerator 0.0027 over denominator k end fraction end root equals square root of fraction numerator 0.0027 over denominator 10.8 end fraction end root blank equals 0.05 m

    Two rectangular blocks A blankand B blankof masses 2kg and 3 kg respectively are connected by spring of spring constant 10.8 N m to the power of negative 1 end exponentand are placed on a frictionless horizontal surface. The block A blankwas given an initial velocity of 0.15 m s to the power of negative 1 end exponent 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 m s to the power of negative 1 end exponent, 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 m s to the power of negative 1 end exponent. 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