Simply, pull or push is called force. A book lying on the table continues to lie on the table unless an external force is applied on it. So, generally force is an external agency which changes or tends to change the state of a body from rest to motion or vice versa. The SI unit of force is newton (N). In CGS system, the unit of the force is dyne. 1 newton= 10^{5} dynes.
Consider a situation in which football is rolling in the ground. Ball stops rolling after some time. If a ground is perfectly smooth then ball will continue rolling forever in same direction with constant speed. It is a roughness of the surface that provides opposing force, which is called friction, to slow down the speed of the ball and finally stop the ball. Friction can be defined as the opposing force which acts on the opposite direction of motion.
Force acting on a body can do following things:
When a number of forces acting on a body do not change its states of rest or uniform motion in a straight line, the force are said to be balanced forces. For example, in the game tug of war, a game where two teams pull a rope in opposite direction. In this game rope remains in steady and does not move in any direction when two teams exert equal and opposite force in a rope.
When a number of forces acting on a body change its state of rest or uniform motion in a straight line, the forces is said to be unbalanced forces.
For example, if we push a toy car towards east, it moves towards east. This unbalanced force produces motion.
Force is defined as the push or pull on a body which changes or tends to change the state of rest or uniform motion in a straight line. Its SI unit is Newton and CGS unit is dyne.
Force can affect the body in the following ways:
When a number of forces acting on a body changes its state of rest or uniform motion in a straight line, then the forces are said to be unbalanced forces.
Solution:
Given,
Retarding force (f) = -10N
Mass (m) = 20kg
Initial velocity (u) = 10m/s
Time (t) = ?
Final velocity (v) = ?
By formula,
F = ma
or, a = F/m
or, a = -10/20 = -0.5 m/s^{2}
And,
a =
or, t =
or, t = or, t = 20s Hence, the body takes 20 s to stop.
When a number of forces acting on a body do not change its states of rest or uniform motion in a straight line, the force are said to be balanced forces.
Though action and reaction are equal in magnitude and opposite in direction, they act on different bodies. Thus the forces acting on different bodies don't cancel each other. Hence, as they act on different bodies, action and reaction don't cancel each other.
A) Let d the distance between his home and school in a straight line.The time taken by the boy to run from home to school (t1) = \(\frac{d}{3}\)
(speed= \(\frac{distance}{time}\))
The time taken by him to walk from school to home (t2) = \(\frac{d}{5}\)
Total time taken (t) = t1+t2
= \(\frac{d}{3}\) + \(\frac{d}{5}\) = \(\frac{8}{15}\) d
Total Distance Travelled = d+d =2d
Average speed = \(\frac{Total Distance}{Total Time}\)
= \(\frac{2d}{8d}\) x 15
= 3.75 m/s
So,Average speed = 3.75 m/s
Again,
Since the boy comes back to the starting point (i.e.,home) his total displacement is zero.
Average velocity = \(\frac{Total Displacement}{Total Time}\)
= \(\frac{s}{t}\) = \(\frac{0}{8}\) = 0
Average velocity = 0m/s
Total distance covered (s) = 400 km = 3x = \(10^{5}\) m [1km=1000m]
Total time taken (t) = 8 hours = 8 x 3600 s [ 1hr=3600sec]
We know that,
Average speed (V) = = \(\frac{Total\;distance\;covered (S) }{Total\;Time\;Taken(T) }\)
= \(\frac{4 X 10{^5}} {8 X 3600}\)
= 13.89 m/s
Therefore,the average speed of the bus is 13.89 m/s.
Here, initial velocity (u) = 6m/s
Final Velocity = 8m/s
Time Taken(t) = 2s
Acceleration(a) = ?
We know that,
a= \(\frac{v-u}{t}\)
= \(\frac{8-6}{2}\)
= \(\frac{2}{2}\) = 1 m/\(s^{2}\)
Therefore, the acceleration of a car is 1 m/\(s^{2}\)
Here, Initial velocity (u) = 0
Acceleration (a) = \(0.5 m/s^{2}\)
Time Taken (t) = 2 minutes = 120 s [1 min=60 sec]
Final Velocity (V) = ?
Distance Covered (S)= ?
We have,
V=u+at
or, v = 0+0.5 x 120
v= 60m/s
Again,
s= \(\frac{u+v}{2}\)
= \(\frac{0+60}{2}\) x 120
= 3600m
Therefore, the final veocity of the bus is 60m/s and it will cover 3600m in 2 minutes
Here, Initial Velocity (u) = 40m/s
For upwoard direction, acceleration (a) = -10m/\(s^{2}\)
The velocity of the ball goes on decreasing as the ball attains height. At the highest point, final velocity (v)=0
We have,
v=u+at
or,0=40+(-10)t
or 10t= 40
or t = \(\frac{40}{10}\)
t= 4s
Again, we have
s= ut+\(\frac{1}{2}\) \(at^{2}\)
or,s= 40 x 2 + \(\frac{1}{2}\) (-10) (\(2^{2}\))
s= 80 + \(\frac{1}{2}\) x (-40)
or, s = 80-20 = 60 m
Since 2s is the time taken by the ball to reach the maximum height, so time taken to return to the initial position is 2t= 2x2 = 4s
Therefore, the maximum height attained is 60 m and the time taken for it to return to the initial position is 4s.
Although the driver sees a baby 60 m ahead on the road , he stops the car at a distance of 15 m, hence,
Distance covered (s) = 15m
Initial Velocity (v) = 0
90km/hr = \(\frac {60 \times 1000m }{60 \times 60s }\)
= 16.66 m/s
Retardation = ?
Time taken (t) = ?
We know ,
\(v^{2}\) =\(u^{2}\) + 2as
or, \(0^{2}\) = ( \(16.66^{2}\) ) + 2a\(\times\) 15
or, -625 = 30a
or, a = \(\frac{277.5}{30}\)
\(\therefore\) a= -9.25 m/\(s^{2}\)
Again , we have
v= u + at
or, 0 = 16.66 + (-20.83) t
or, -16.66 = -9.25t
or, t = \(\frac{16.66}{9.25}\)
\(\therefore\) t= 1.80 s
Therefore, the acceleration of the car is -9.25 m/\(s^{2}\). i.e. its retardation 9.25 m/\(s^{2}\) and it takes 1.8 to come at rest.
Here Mass (m) = 100g = \(\frac{100}{1000}\) kg = 0.1
Initial Velocity (u) = 15m/s
Final Velocity (v) = 0
Time Taken (t) = 0.05s
We have,
v= u + at
or, 0 = 15 + a \(\times\) 0.05
or, a = \(\frac{15}{0.05}\)
\(\therefore\) a = -300 m/\(s^{2}\)
Again, we have
F= ma
or,F= 0.1 \(\times\) (-300)
\(\therefore\) = -30N
Here,
Mass (m) = 100g = \(\frac{100}{1000}\) kg
= 0.1kg
Distance Covered (s)= 200cm = 2m
Time Taken (t) = 5s
Initial Velocity (u) = 0
We know,
\(\therefore\) s= ut + \(\frac{1}{2}\) \(at^{2}\)
or, 2= 0 \(\times\) t + \(\frac{1}{2}\) a \( (5)^{2}\)
a= 2 \(\frac{4}{25}\) m/ \(s^{2}\)
Again, we have
\(\therefore\) F = ma
or, F= (0.1) \(\times\) \(\frac{4}{25}\) = 0.016 N
Thus, the magnitude of the force is 0.016 N
Solution:
Here, Initial Velocity (U) = 0
Acceleration(a) = 0.5m/s \(^2\)
Time taken (t) = 2 minutes = 120 s
Final velocity (V) = ?
Distance Covered(s) = ?
We have,
v = u + at
or, v = 0 + 0.5 \(\times\) 120
v = 60m/s
Again,
s = \(\frac{u\;+v}{2}\) \(\times\) t
= \(\frac{o\;+60}{2}\) \(\times\) 120 = 3600 m
Therefore, the final velocity of the bus is 60m/s and it will cover 3600 m in 2 minutes.
Here, Initial velocity (u) = 20m/s
Acceleration (a) = -10m/s \(^2\)
Final Velocity (v) = 0
We have,
v = u + at
or, 0 = 20 + (-10) t
or, 10t = \(\frac{20}{10}\)
\(\therefore\) t = 2s
Again, we have
s = ut + \(\frac{1}{2}\) \(at^{2}\)
or, s = 20 \(\times\) 2 + \(\frac{1}{2}\) (-10) (2^{2)}or, s = 40 + \(\frac{1}{2}\) \(\times\) (-40)
or, s = 40 - 20 = 20m
Since, 2s is the time taken by the ball to reach the maximum height, so time taken to return to the initial position is 2t = 2 \(\times\) 2s = 4s.
Therefore, the maximum height attained is 20 m and the time taken is 20m and the time taken for it to return to the initial position is 4s.
Force is measured using ______.
beam balance
newton meter
pan balance
accerelometer
________ act on an object from opposite sides, but make the object move.
Unbalanced force
Balance force
Motion
Pressure
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