General
Easy
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

Physics-General

  1. 225 J    
  2. 200 J    
  3. 400 J    
  4. 175 J    

    Answer:The correct answer is: 200 JWork 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

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    The potential energy of a system is represented in the first figure. The force acting on the system will be represented by

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    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

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    As slope of problem graph is positive and constant upto certain distance and then it becomes zero
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    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

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    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.
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    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
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    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?

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    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
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    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
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    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

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

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    If the surface is smooth then the kinetic energy at B never be zero
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    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

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    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
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    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
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    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
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