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A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. The speed is then increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks.
A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform speed of 10 m s−1, what would be the reading on the scale?
Mass of the man, m = 70 kg
Acceleration, a = 0
Using Newton’s second law of motion, We can write the equation of motion as,
R – mg = ma
∴ R = mg = 70 × 10 = 700 N
∴ the weighing scale = 700 / g = 700 / 10 = 70 kg
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg lying on the floor of a train which is accelerating with 1 ms^{−2}, the stone being at rest relative to the train. Neglect air resistance.
As the stone is lying on the floor of the train, its acceleration is same as that of the train.
∴ Force acting on stone,
F = ma = 0.1 x 1 = 0.1 N.
According to Newton's third law of motion, the action and reaction forces are
If net force act on a body then acceleration of the body will also be zero. Hence velocity will not be changed i.e. it continues in its existing state of rest or uniform motion in a straight line.
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg lying on the floor of a train which is accelerating with 1 ms^{−2}, the stone being at rest relative to the train. Neglect air resistance.
Weight of the stone is balanced by the reaction of the floor. The only acceleration is provided by the horizontal motion of the train.
a = 1ms^{2}
Force in horizontal direction
F = ma = 0.1 x 1 = 0.1N
A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 10 ms^{−2} what would be the reading on the scale?
According to Newton's second Law of motion,
Apparent weight, R = m(a+g)
= 70(10+10)
= 1400N
mass = R/g
= 1400/10
= 140Kg
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms^{−2}. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the force on the floor by the crew and passengers,
A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 106 kg. What is the angle of banking required to prevent wearing out of the rail?
Radius of the circular track, r = 30 m
Speed of the train, v = 54 km/h = 15 m/s
Mass of the train, m = 106 kg
The centripetal force is provided by the lateral thrust of the rail on the wheel.
As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail.
The angle of banking θ, is related to the radius (r) and speed (v) by the relation:
tan θ = v^{2} / rg
= 152 / (30 x 10)
= 225 / 300
⇒ θ = tan^{1} (0.75) = 36.870
Therefore, the angle of banking is about 36.87degree = 37degree.
When no external forces act on a system of several interacting particles, then the total momentum of system is conserved.
Law of conservation of Momentum:
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s^{−1}How long does the body take to stop?
According to Newton's second Law
Force = mass x accleration
F = m.a
= m.v/t
t = m.v/F
= 20x15/50
= 6 sec
A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 ms^{−2} what would be the reading on the scale?
Mass of the man, m = 70 kg
Acceleration, a = 5 m/s2 upward
Using Newton’s second law of motion, we can write the equation of motion as:
R – mg = ma
R = m(g + a)
= 70 (10 + 5) = 70 x 15
= 1050 N
∴ Reading on the weighing scale = 1050 / g = 1050 / 10 = 105 kg
A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. What is the tension in the rope if the monkey climbs up with an acceleration of 6 ms^{−2}
Mass of the monkey, m = 40 kg
Acceleration due to gravity, g = 10 m/s
Maximum tension that the rope can bear, Tmax = 600 N
Acceleration of the monkey, a = 6 m/s2 upward
Using Newtons second law of motion, we can write the equation of motion as:
T mg = ma
T = m(g + a)
= 40 (10 + 6)
= 640 N
Newton's second law of motion gives the quantitative definition of force. The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
A rocket with a liftoff mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms^{2}. Calculate the initial thrust (force) of the blast.
Initial thrust = upthrust required to impart acceleration + uthrust to overcome gravity
⇒ ma + mg = m(a+g)
⇒ 20000(5+10) = 300000 N
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. a horizontal force F = 600 N is applied to the 20 kg mass so as to pull it. What is the tension in the string?
For some instance assume both masses as one system, thus we get that
600 = 30a
Where a is the common acceleration of the system.
Now if we consider the 10kg block we get that
T = 10a
And a = 20m/s^{2 }
Thus we get T = 200N
A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. What is the tension in the rope if the monkey climbs down with an acceleration of 4 ms^{−2}
Mass of the monkey, m = 40 kg
Acceleration due to gravity, g = 10 m/s
Maximum tension that the rope can bear, Tmax = 600 N
Acceleration of the monkey, a = 4 m/s2 downward
Using Newton’s second law of motion, we can write the equation of motion as:
mg T = ma
T = m (g a)
= 40(104)
= 240 N
When an object has several forces acting on it, the effect of force is the same as one force acting on the object in a certain direction and this overall force is called the ‘resultant force’. The resultant force is essential to change the velocity of an object.
A truck starts from rest and accelerates uniformly at 2.0 ms^{−2}. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What is the magnitude of acceleration (in ms^{−2} ) of the stone at t = 11s? (Neglect air resistance.)
Initial velocity of the truck, u = 0
Acceleration, a = 2 m/s2
Time, t = 10 s
As per the first equation of motion, final velocity is given as:
v = u + at
⇒ 0 + 2 X 10 = 20 m/s
The final velocity of the truck and hence, of the stone is 20 m/s.
At t = 11 s, the horizontal component (v_{x}) of velocity, in the absence of air resistance, remains unchanged, i.e.,
v_{x} = 20 m/s
The vertical component (v_{y}) of velocity of the stone is given by the first equation of motion as:
v_{y} = u + a_{y}δt
Where, δt = 11 – 10 = 1 s and a_{y} = g = 10 m/s2
∴v_{y} = 0 + 10 X 1 = 10 m/s
A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. What is the tension in the rope if the monkey climbs up with a uniform speed of 5 ms^{−1}
When monkey climbs with uniform speed, the acceleration is zero, so
T = mg = 40x10 = 400 N
A stationary ball weighing 0.25kg acquires a speed of 10m/s when hit by a hockey stick. The impulse imparted to the ball is
Initial momentum of the ball,
P_{i} = 0
Final momentum of the ball,
P_{f} = mv
⇒ 0.25×10 = 2.5 kg m/s
Impulse imparted on the ball,
I = P_{f} − P_{i}
⇒ I = 2.5 − 0 = 2.5 N s
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms^{−1}. What is the trajectory of the bob if the string is cut when the bob is at one of its extreme positions
At the extreme position of the oscillation, the speed of the bob is zero. So the bob is momentarily at rest. if the string is cut, the bob will fall vertically downwards.
The net external force on the rigid body is always equal to the total mass times the translational acceleration (i.e., Newton's second law holds for the translational motion, even when the net external torque is nonzero, and/or the body rotates).
Rolling friction is always less than static friction because in order for an object to roll the force of friction between it and surface must be large enough to keep the object from sliding. Hence rolling friction is always greater than static force.
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