Class 12 Exam > Class 12 Videos > L20 : Problem law of motion - Laws of motion, Physics, Class 11
L20 : Problem law of motion - Laws of motion, Physics, Class 11 Video Lecture - Class 12
FAQs on L20 : Problem law of motion - Laws of motion, Physics, Class 11 Video Lecture - Class 12
| 1. What are the three laws of motion? |  |
Ans. The three laws of motion, also known as Newton's laws of motion, are as follows:
1. Newton's First Law of Motion: An object at rest will remain at rest, and an object in motion will continue to move at a constant velocity unless acted upon by an external force.
2. Newton's Second Law of Motion: The acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. This is described by the equation F = ma, where F is the net force, m is the mass, and a is the acceleration.
3. Newton's Third Law of Motion: For every action, there is an equal and opposite reaction. This means that if an object exerts a force on another object, the second object will exert an equal but opposite force on the first object.
| 2. How do Newton's laws of motion apply to everyday life? | |
Ans. Newton's laws of motion apply to everyday life in various ways:
- Newton's First Law: It explains why objects tend to stay at rest or keep moving unless acted upon by a force. For example, when a car suddenly stops, the passengers tend to move forward due to their inertia.
- Newton's Second Law: It helps us understand how the acceleration of an object depends on the force applied and its mass. For instance, pushing a heavy object requires more force to accelerate it compared to a lighter object.
- Newton's Third Law: It explains the concept of action and reaction. For example, when we walk, our feet push against the ground, and the ground pushes back with an equal force, propelling us forward.
| 3. How does Newton's first law of motion relate to seatbelts in cars? | |
Ans. Newton's first law of motion, also known as the law of inertia, relates to seatbelts in cars as follows:
When a car suddenly stops or changes its speed, the passengers inside the car tend to continue moving at their original velocity due to their inertia. If unrestrained, this can cause them to be thrown forward and potentially collide with the car's interior or be ejected from the vehicle in case of an accident.
Seatbelts help protect passengers by providing a restraining force that keeps them in place during sudden deceleration. By wearing seatbelts, the force exerted on the passengers' bodies is spread out over a larger area, reducing the risk of injury.
| 4. How is force related to acceleration according to Newton's second law of motion? | |
Ans. According to Newton's second law of motion, the force acting on an object is directly proportional to its acceleration. This law is mathematically represented by the equation F = ma, where F is the net force applied to the object, m is the mass of the object, and a is the acceleration it experiences.
In simpler terms, the greater the force applied to an object, the greater its acceleration will be. Similarly, if the mass of the object increases, the force required to achieve a certain acceleration also increases.
| 5. Can you provide an example of Newton's third law of motion? | |
Ans. One example of Newton's third law of motion is the recoil of a firearm. When a bullet is fired from a gun, the gun experiences a backward force, known as recoil, as a reaction to the forward force exerted by the bullet. The bullet is propelled forward due to the force exerted by the expanding gases in the gunpowder, and as a result, the gun moves backward. This is an illustration of how every action has an equal and opposite reaction, as stated in Newton's third law of motion.
Text Transcript from Video
hello friends this video on laws of
motion part 20 is brought to you by
example.com no more your prom exam
please make sure that you have watched
all the visuals from part 1 to 19 before
going ahead with part 20 the problem
cells and in extensible string connects
blocks m1 is equal to 10 kgs and m2 is
equal to 5 kg passing over a light
frictionless pulley so what does it say
it connects two blocks m1 and m2 over a
frictionless pulley okay
m1 moves over a frictionless surface by
m2 moves vertically so let us draw the
diagram first so let us suppose this is
m1 the question says m1 moves over a
frictionless surface so let us say this
is the frictionless surface it also says
m1 and m2 are connected through a
frictionless pulley and m2 moves
vertically that means they must be
connected somewhat like this let us
suppose this is the pulley and let us
say this is m2 so m2 is moving
vertically and m1 will be moving over
the frictionless surface now find the
acceleration of the system and the
tension in the string now let us draw
the free body diagram for both the mass
separating which will make it easy for
you to understand let us suppose for m2
the weight will be acting downwards that
is m2 G tension will be acting in this
direction since it is a frictionless
pulley so tension along this and tension
along this will be the same here is not
a tension T which is in this direction
and now we will write the equation of
motion for both the masses now let us
suppose that this mass m2 moves in the
downward direction and the acceleration
for this is a 2 fine
similarly let us suppose the
acceleration for this mass is a 1 let us
say that our motion is in this direction
now since it is a frictionless pulley
therefore a 1 will be equal to a 2 so
let us say a 1 a 2 is equal to K so let
us consider this acceleration is a this
acceleration is also a now we will write
the equation of motion for both the
masses let us write the equation for m2
first so what will be the equation for M
2 the net force on M 2 will be equal to
the force which is equal to mass into
acceleration so what will be the net
force here m 2g minus T is equal to then
force what is force forces mass into
acceleration so it will be mass into
acceleration so this will be the
equation for mass M 2 so what we
basically do to write the equation the
net force acting on this mass so what is
the net force acting here this M 2 G and
T now since it is moving in the downward
direction so we say that m2 G minus T is
the net force and what is the force with
which it is moving that is mass into
acceleration
you already know force is mass into
acceleration so this is m2 G minus T is
equal to Ma now let us write a similar
equation for M 1 now what is the net
force on M 1 T there is no other force
which is acting it moreover there is no
friction so there will be no frictional
force is it or the question already seen
that it is a frictionless surface that
means there will be no frictional force
between the block and the surface so the
net force on this is T so this T will
again become equal to force that is mass
into acceleration so that will be M 1
into n I am sorry here mr. this was M 2
because the mass of this block is empty
so this is my second equation now what
does the question asks the question asks
us to find the value of a this a we have
to find out right so
see how we are able to find it out we
got two equations now let us add up the
two equations so what do we get on our
left hand side we get M - gene - t plus
T is equal to M 2 a plus M 1 K so minus
T plus T gets cancelled so we get M 2 G
is equal to a into M 1 plus M 2 now I
feel that you must be finding it pretty
simple to calculate a so it is M 2 G
divided by M 1 plus M 2 so see friends
whenever you have any problems related
to pulley blobs or something like that
the best way to solve it is you first
draw the free body diagram for both the
bodies and you write the equation of
motion whenever you write equation of
motion just try to find the net force on
that block like here the net force is m2
G minus T so this net force has to be
equal to mass into acceleration
similarly here the net force is T so the
net force has to be equal to mass into
acceleration so this is the Panther
which you need to follow to write the
equation of motion so I write it here to
write equation of motion you just
remember net force on the body is equal
to mass into acceleration just remember
this now once you are able to draw the
free body diagram correctly it will be
easier it is very easy for you to find
out the net force on the body so here we
where acceleration is m2 G divided by m1
plus m2 now in the question we are
already given that m1 is ten cages and
m2 is 5 kgs so let us put the values M 2
is 5 G is 9.8
m1 is 10 M 2 is again five so we get
five into 9.8
/ 15 so we get the value as 3.27 meter
per second square so this is the value
of the acceleration of the system now
the question asks find the acceleration
of the system and the tension in the
string now once you have got the value
of a it is very easy for you to
calculate the value of T so from
Equation 2 you can say T is equal to m1
into a so what is M 1 M 1 is 10kgs into
3.27
so it comes out to be 32 point 7 Newton
so this is the tension I hope this is
clear to you now since we are solving
many different kinds of problems so I
guess the concept is getting clearer to
you as we proceed further let's go ahead
and see more examples this problem says
a mass of four kg rests on a horizontal
plane the plane is gradually inclined
until at an angle theta is equal to 15
degree with the horizontal so let us
imagine the scenario practically let us
suppose this was of horizontal plane on
which the four kg mass was resting now
we started inclining this plane that
means gradually we made this plane
somewhat like this I mean we inclined it
we kept inclining it and finally we made
it inclined to an angle of 15 degree
with our is on now if you think of the
situation practically then what will
happen to this block as soon as we start
in planning this plane the block will
start sliding correct now you have to
find the coefficient of static friction
between the block and the surface so
block coefficient of static friction
that means this static friction will be
there mind at what point the box will
actually start sliding right so let us
see how we can solve this first of all
we will draw the diagram let us say this
is our inclined plane this is the mass
now let us draw all the forces
that will be acting on this mask now
this angle is Theta and in the question
it is given that theta is equal to 15
degree and also mass of this block is
equal to 4 kgs now what are the
different forces that will act on this
block obviously mg will act downward
this thing normal reaction will act
normal to the surface of contact so this
is my surface of contact so it will act
normal to this surface that means it
will act in this direction so this is my
normal reaction okay now since this
angle is and what else will act on this
block now if the block will slide in
this direction whenever it slides it
will slide in this direction right so
the force of the static frictional force
will act definitely along this direction
so this will be the direction of static
friction that is FS now let us find the
components of mg now what we have to do
we will write the equation we will write
the equation of motion along F as well
as y-axis one by one like so let us
first see the equation let us first try
to find the components of mg now if this
is mg this is Theta this angle will also
become theta so if this is mg this is
theta so what will be this this will be
mg cos theta and what will be this this
would be mg sine theta right now it is
very simple for you to find the equation
so along let us consider that this is my
x-axis let us consider this is my x axis
and this is my y axis of coordinate
system okay so along x axis that is
along the horizontal direction what will
be the equation of motion it would be if
s is equal to mg sine theta correct
because we want to find out the
címon value I mean we want to find out
the coefficient of static friction
now once the static friction vanishes
that is once the static friction reaches
its maximum value this box will start
sliding
so the question talks about - scenario
when the box just begins to slide that
means at that point of time the value of
FS will be the maximum value of F this
correct
so along x axis let's first equate it is
FS is equal to mg sine theta along y
axis it would be n is equal to mg cos
theta because it the masses please keep
in mind friends now the masses just
begins to slide that means till now the
mass was at rest so there is no motion
that is why the force along this
direction will be equal to the force
along this direction right so we found
two equations now according to the
question if say is that the Box just
begin to slide just begin to slide so
this says that the value of F F is the
maximum value of FS which is equal to
what is our maximum value of F is it is
mu s into n this is the maximum value of
FS so we will put this value in equation
one so what do we get we get mus n is
equal to mg sine theta so equation one
becomes that is mu s n is equal to mg
sine theta now we can substitute the
value of n as mg cos theta from this
equation so we get mu s mg cos theta is
equal to mg sine theta so mg mg gets
cancelled
so what do we get we get mu s is equal
to sine theta by cos theta which comes
out to be tan 15-degree so your answer
will be the value of coefficient
static friction is 1015 degree so I hope
it is clear now here also we did the
same thing we first we drew the diagram
we find out all the forces acting on the
body since the body is in rest therefore
the all the forces which are acting
along horizontal direction they should
be equal and opposite
similarly the sum of all the forces
which is acting along the vertical
direction should also be equal to zero
that is these forces should be equal and
opposite right so we wrote both the
equations since this body just begun to
slide therefore the value of frictional
force that is the static frictional
force was its maximum value or limiting
value which is mu s into n so we just
substituted the value and from these two
equations we were able to find out the
value of mu s so I hope now things are
getting clearer to you
so whenever the body is moving in that
case we write equation of motion as the
net force acting on the body is equal to
mass into acceleration when a body is at
rest in that case the sum of all the
forces acting on the body should be
equal to zero thank you visits exam
viacom to watch free educational videos
try3 online tests get the best quality
study materials study from the best
tutors and mentors and much more thanks
once again
motion part 20 is brought to you by
example.com no more your prom exam
please make sure that you have watched
all the visuals from part 1 to 19 before
going ahead with part 20 the problem
cells and in extensible string connects
blocks m1 is equal to 10 kgs and m2 is
equal to 5 kg passing over a light
frictionless pulley so what does it say
it connects two blocks m1 and m2 over a
frictionless pulley okay
m1 moves over a frictionless surface by
m2 moves vertically so let us draw the
diagram first so let us suppose this is
m1 the question says m1 moves over a
frictionless surface so let us say this
is the frictionless surface it also says
m1 and m2 are connected through a
frictionless pulley and m2 moves
vertically that means they must be
connected somewhat like this let us
suppose this is the pulley and let us
say this is m2 so m2 is moving
vertically and m1 will be moving over
the frictionless surface now find the
acceleration of the system and the
tension in the string now let us draw
the free body diagram for both the mass
separating which will make it easy for
you to understand let us suppose for m2
the weight will be acting downwards that
is m2 G tension will be acting in this
direction since it is a frictionless
pulley so tension along this and tension
along this will be the same here is not
a tension T which is in this direction
and now we will write the equation of
motion for both the masses now let us
suppose that this mass m2 moves in the
downward direction and the acceleration
for this is a 2 fine
similarly let us suppose the
acceleration for this mass is a 1 let us
say that our motion is in this direction
now since it is a frictionless pulley
therefore a 1 will be equal to a 2 so
let us say a 1 a 2 is equal to K so let
us consider this acceleration is a this
acceleration is also a now we will write
the equation of motion for both the
masses let us write the equation for m2
first so what will be the equation for M
2 the net force on M 2 will be equal to
the force which is equal to mass into
acceleration so what will be the net
force here m 2g minus T is equal to then
force what is force forces mass into
acceleration so it will be mass into
acceleration so this will be the
equation for mass M 2 so what we
basically do to write the equation the
net force acting on this mass so what is
the net force acting here this M 2 G and
T now since it is moving in the downward
direction so we say that m2 G minus T is
the net force and what is the force with
which it is moving that is mass into
acceleration
you already know force is mass into
acceleration so this is m2 G minus T is
equal to Ma now let us write a similar
equation for M 1 now what is the net
force on M 1 T there is no other force
which is acting it moreover there is no
friction so there will be no frictional
force is it or the question already seen
that it is a frictionless surface that
means there will be no frictional force
between the block and the surface so the
net force on this is T so this T will
again become equal to force that is mass
into acceleration so that will be M 1
into n I am sorry here mr. this was M 2
because the mass of this block is empty
so this is my second equation now what
does the question asks the question asks
us to find the value of a this a we have
to find out right so
see how we are able to find it out we
got two equations now let us add up the
two equations so what do we get on our
left hand side we get M - gene - t plus
T is equal to M 2 a plus M 1 K so minus
T plus T gets cancelled so we get M 2 G
is equal to a into M 1 plus M 2 now I
feel that you must be finding it pretty
simple to calculate a so it is M 2 G
divided by M 1 plus M 2 so see friends
whenever you have any problems related
to pulley blobs or something like that
the best way to solve it is you first
draw the free body diagram for both the
bodies and you write the equation of
motion whenever you write equation of
motion just try to find the net force on
that block like here the net force is m2
G minus T so this net force has to be
equal to mass into acceleration
similarly here the net force is T so the
net force has to be equal to mass into
acceleration so this is the Panther
which you need to follow to write the
equation of motion so I write it here to
write equation of motion you just
remember net force on the body is equal
to mass into acceleration just remember
this now once you are able to draw the
free body diagram correctly it will be
easier it is very easy for you to find
out the net force on the body so here we
where acceleration is m2 G divided by m1
plus m2 now in the question we are
already given that m1 is ten cages and
m2 is 5 kgs so let us put the values M 2
is 5 G is 9.8
m1 is 10 M 2 is again five so we get
five into 9.8
/ 15 so we get the value as 3.27 meter
per second square so this is the value
of the acceleration of the system now
the question asks find the acceleration
of the system and the tension in the
string now once you have got the value
of a it is very easy for you to
calculate the value of T so from
Equation 2 you can say T is equal to m1
into a so what is M 1 M 1 is 10kgs into
3.27
so it comes out to be 32 point 7 Newton
so this is the tension I hope this is
clear to you now since we are solving
many different kinds of problems so I
guess the concept is getting clearer to
you as we proceed further let's go ahead
and see more examples this problem says
a mass of four kg rests on a horizontal
plane the plane is gradually inclined
until at an angle theta is equal to 15
degree with the horizontal so let us
imagine the scenario practically let us
suppose this was of horizontal plane on
which the four kg mass was resting now
we started inclining this plane that
means gradually we made this plane
somewhat like this I mean we inclined it
we kept inclining it and finally we made
it inclined to an angle of 15 degree
with our is on now if you think of the
situation practically then what will
happen to this block as soon as we start
in planning this plane the block will
start sliding correct now you have to
find the coefficient of static friction
between the block and the surface so
block coefficient of static friction
that means this static friction will be
there mind at what point the box will
actually start sliding right so let us
see how we can solve this first of all
we will draw the diagram let us say this
is our inclined plane this is the mass
now let us draw all the forces
that will be acting on this mask now
this angle is Theta and in the question
it is given that theta is equal to 15
degree and also mass of this block is
equal to 4 kgs now what are the
different forces that will act on this
block obviously mg will act downward
this thing normal reaction will act
normal to the surface of contact so this
is my surface of contact so it will act
normal to this surface that means it
will act in this direction so this is my
normal reaction okay now since this
angle is and what else will act on this
block now if the block will slide in
this direction whenever it slides it
will slide in this direction right so
the force of the static frictional force
will act definitely along this direction
so this will be the direction of static
friction that is FS now let us find the
components of mg now what we have to do
we will write the equation we will write
the equation of motion along F as well
as y-axis one by one like so let us
first see the equation let us first try
to find the components of mg now if this
is mg this is Theta this angle will also
become theta so if this is mg this is
theta so what will be this this will be
mg cos theta and what will be this this
would be mg sine theta right now it is
very simple for you to find the equation
so along let us consider that this is my
x-axis let us consider this is my x axis
and this is my y axis of coordinate
system okay so along x axis that is
along the horizontal direction what will
be the equation of motion it would be if
s is equal to mg sine theta correct
because we want to find out the
címon value I mean we want to find out
the coefficient of static friction
now once the static friction vanishes
that is once the static friction reaches
its maximum value this box will start
sliding
so the question talks about - scenario
when the box just begins to slide that
means at that point of time the value of
FS will be the maximum value of F this
correct
so along x axis let's first equate it is
FS is equal to mg sine theta along y
axis it would be n is equal to mg cos
theta because it the masses please keep
in mind friends now the masses just
begins to slide that means till now the
mass was at rest so there is no motion
that is why the force along this
direction will be equal to the force
along this direction right so we found
two equations now according to the
question if say is that the Box just
begin to slide just begin to slide so
this says that the value of F F is the
maximum value of FS which is equal to
what is our maximum value of F is it is
mu s into n this is the maximum value of
FS so we will put this value in equation
one so what do we get we get mus n is
equal to mg sine theta so equation one
becomes that is mu s n is equal to mg
sine theta now we can substitute the
value of n as mg cos theta from this
equation so we get mu s mg cos theta is
equal to mg sine theta so mg mg gets
cancelled
so what do we get we get mu s is equal
to sine theta by cos theta which comes
out to be tan 15-degree so your answer
will be the value of coefficient
static friction is 1015 degree so I hope
it is clear now here also we did the
same thing we first we drew the diagram
we find out all the forces acting on the
body since the body is in rest therefore
the all the forces which are acting
along horizontal direction they should
be equal and opposite
similarly the sum of all the forces
which is acting along the vertical
direction should also be equal to zero
that is these forces should be equal and
opposite right so we wrote both the
equations since this body just begun to
slide therefore the value of frictional
force that is the static frictional
force was its maximum value or limiting
value which is mu s into n so we just
substituted the value and from these two
equations we were able to find out the
value of mu s so I hope now things are
getting clearer to you
so whenever the body is moving in that
case we write equation of motion as the
net force acting on the body is equal to
mass into acceleration when a body is at
rest in that case the sum of all the
forces acting on the body should be
equal to zero thank you visits exam
viacom to watch free educational videos
try3 online tests get the best quality
study materials study from the best
tutors and mentors and much more thanks
once again
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Dec 25, 2024 Last updated |
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