A block of mass m=25 kg sliding on a smooth horizontal surface -Turito

General

Easy

Physics-

A block of mass $m equals 25$ kg 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

$1.5 m, - 3 {m s}^{- 1}$
$1.5 m, 0.01 {m s}^{- 1}$
$1.0 m, 3 {m s}^{- 1}$
$0.5 m, 2 {m s}^{- 1}$

Answer:The correct answer is: $1.5 blank m comma blank minus 3 blank m s to the power of negative 1 end exponent$ 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.
$H e n c e v equals negative 3 m s to the power of negative 1 end exponent$

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Related Questions to study

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 $x equals 1 m$ to $x equals 5$ m will be

Work done=area enclosed by

F minus x

graph
=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 $x equals 1 m$ to $x equals 5$ m will be

physics-General

Work done=area enclosed by

F minus x

graph
=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-

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-

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

Work 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

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

Work 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

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-

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-

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 $x equals 8.0$ m?

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

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 $x equals 8.0$ m?

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

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-

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 $12 blank m$

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Can’t find what you’re looking for?

$1.5 m, - 3 {m s}^{- 1}$
$1.5 m, 0.01 {m s}^{- 1}$
$1.0 m, 3 {m s}^{- 1}$
$0.5 m, 2 {m s}^{- 1}$

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Related Questions to study

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 $x equals 1 m$ to $x equals 5$ m will be

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 $x equals 1 m$ to $x equals 5$ m will be

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

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

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

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

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

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

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 $x equals 8.0$ m?

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 $x equals 8.0$ m?

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 $12 blank m$

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 $12 blank m$

Have a query? Ask a Tutor!!

Can’t find what you’re looking for?

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

Book A Free Demo

Related Questions to study

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

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

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

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

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

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

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

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

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?

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?

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

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

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 $x equals 1 m$ to $x equals 5$ m will be

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 $x equals 1 m$ to $x equals 5$ m will be

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

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

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

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

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

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

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 $x equals 8.0$ m?

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 $x equals 8.0$ m?

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 $12 blank m$

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 $12 blank m$