Numerical based on newton's laws of motion
here are following steps to solve any numerical problem-:
step 1:- first of all decide the system and draw its F.B.D. separately from the given diagram.
step 2:- write the name of forces acting on it.
Note:- normal force act perpendicular to the surface of body.normal force acts because of contact between the surface of body.It is represented by N
👉gravity force (mg) alway acts vertically downward to the surface.it acts because of gravitional force of earth. here mass of block M.
👉tension alway act in the thread , wire ,rope .or any rod when it is taut .tension always act in the inward direction. here tension in rope given in diagram.
step :-4 write the equation of all the forces along the perpendicular motion ,and make it =0 i.e. fnet =0. 
here only tension force is acting along the accleration.so....
fnet =ma
T =ma equation -(1)
T =ma equation -(1)
fnet = 0
N - mg =0 ----- equation (2)
step :-5 By eliminating any one of variables find the other which is asked in question.
Note -: along acceleration direction of force is taken as +ve.
opposite to the acceleration direction of force is taken as-ve
forces which is acting in +ve y axis sign of force is taken as +ve
force which is acting in -ve y axis sign of force is taken as -ve.
question -: (1) A block of mass 10 kg is resting on a horizontal smooth surface. A man apply the force of 10 N then what will be the acceleration of this block ?
Sol -: F.B.D. of the block of mass 10kg
Along acceleration
Fnet = ma
10 = 10 *a
⇒10/10 = a
⇒1 = a
ஃ acceleration = 10m/s*s . ans .
question (2) -: find the acceleration of both the block and also find the contact force between the blocks . assume all surfaces are smooth.
F.B.D. of block of mass m1
Along acceleration
Fnet = ma
F- N3 = m1a ---------- equation (1)
Along perpendicular motion
Fnet = o
N1- m1g = 0
N1 = m1g --------- equation (2)
Draw the F.B.D of block of mass m2
Along perpendicular motion
Fnet = 0
N2 - m2*g = 0
N2 = m2*g --------- equation (3)
Along accleration
Fnet = ma
N3 = m2a ---------- equation (4)
putting the value of N3 in the equation (1)
F -m2a = m1a
F = m1a+m2a
F = a(m1+m2)
F/(m1+m2) = a
ஃ Acceleration of both block taken as common considering both as single system. = F/ (m1+m2)
n0w , putting the value of acceleration in equation (4)
N3 = m2*F/(m1+m2) ans.
question (3) -: Two blocks of mass 50 kg and 1 kg resting on each other on a smooth horizontal surface . 10 N of force is applied on the block 1 kg . find the acceleration of both blocks and normal forces ?
Draw the f.b.d. of 1 kg block 
Along perpendicular motion
Fnet = 0
N1 - (1*g +N2 ) = 0
N1 = 10+N2 ( g = 10) -------- equation (1)
Along acceleration
Fnet = ma
10 = 1*a
a = 10 m/s*s
 |
F.B.D. of 50 kg block |
Along perpendicular motion
Fnet = 0
⇒ N2 - 50*g = 0
⇒ N2 = 500 N
Force exerted by 1 kg block on the block 50 kg .
N1 = 10 +N2
⇒N1 = 10+ 500
⇒ N1 = 510 N
question (4)-:Three blocks of mass 3 kg 2 kg and 1 kg are placed side by side on a smooth surface as shown in figure . A horizontal force of 12 N is applied on 3 kg block .find the net force on 2 kg block.
Soln-: since all the blocks are moving with same acceleration, so take all the blocks as system.
Draw the free body diagram taking all the block as the system
Along acceleration
Fnet = ma
12 = (3+2+1)a
12/6 = a
2m/s*s = a Ans.
FREE BODY DIAGRAM OF 2KG
Along perpendicular motion
Fnet = 0
N- 2g = 0
N = 2g
Along acceleration
Fnet = ma
F = 2 *2 = 4 N ans.
Net force acting on 2 kg block = 4 N In the arrangement shown in figure . The string are light and inextensible . the surface is over which block is placed smooth . find - (1) The acceleration of each block
(2) The tension in the string
Let the acceleration of all block be a and tension in the string be T1 and T2 . 
Along acceleration
Fnet = ma
14 = (4+2+1)*a
14 = 7a
14/7 = a
2 m/s*s = a
∴ acceleration of all three block = 2m/s*s.
Free body diagram of 4 Kg block
Along acceleration
F net = ma
T1 = 4*2
T1 = 8 N ans.
Along perpendicular motion
F net = 0
N - 4g = 0
N = 4g
Free body diagram of 1 kg block
Along acceleration
Fnet = ma
F- T2 = 1* 2
⇒14- T2 = 2
⇒ 14-2 = T2
⇒12 = T2 Ans.
Note -: jis direction me acceleration hoga us direction ka force bada hoga as compare to force which is acting in the opposite direction of acceleration.
question - ( 6) Two block of mass 4 Kg and 2 kg are attached by an inextensible light string as shown in figure . both the blocks are pulled vertically upward by a force f= 120 N
Find -: (1) the acceleration of blocks
(2) Tension in the string
soln -: Taking the two block and string as a system .free body diagram of this system.
Along acceleration
Fnet = ma
120 - (4g+2g) = (4+2)a
60 = 6a
a = 10 m/s*s . Ans.
Free body diagram of 2 kg block
Along acceleration
Fnet = ma
T - 2g = 2a
T - 20 = 2*10
T = 40 N Ans.
que -: (7) Both the springs shown are unstreched . if the block is displaced by a distance x and released . what will be initial acceleration ?
for the spring block system ----: (1) we find the mean position of the system
mean position is the position where fnet force acting on it is 0. also if the forces is acting is acting then that should be equal and opposite and acting in the opposite direction.
(2) Displace the object from the mean position by distance x.
(3) write the equation force by taking the force positive which is acting in the direction of acceleration and -ve force which is acting in the opposite direction of acceleration.
draw the free body diagram of block m by taking as system.
along perpendicular motion
Fnet = 0
N - mg = 0
N = mg
along acceleration
fnet = ma
k1x +k2x =ma
k1x+k2x /m = a
so acceleration of the block = k1x+k2x/m ans. 3
👉No body stay in rest on any inclined plane it moves at the acceleration of gsinθ.
👉A body stay in rest on any incline plane only if and only when the incline itself moves at the acceleration of gtanθ .
The above free body diagram is made to proof acceleration of ball is 0 only when the incline move with acceleration of g tan theta .
from the above diagram
along acceleration
Fnet = ma
mgsinθ - Fscosθ = m*0
Fscosθ = mgsinθ
macosθ = mgsinθ
acosθ = gsinθ
a = gsinθ /cosθ ⇒ a= gtanθ
A block of mass 2 kg is attached with spring and in hanging state eith the spring of spring constant (k = 100n/m) . find the elongation in the spring ?
free body diagram of 2 kg block.
According to spring force
Fnet = -kx
2g = -100*x
-20/100 = x
-0.2 = x
∴ Elongation of spring = 0.2 m (here - sign show spring force alway act in the opposite direction of displacement of object )
💥 Motion of Elevator
mg = weight of a body in elevator
N = normal reaction of the surface
👉case - 1 when lift is stationary
N = mg
👉case -2 when lift is moving with uniform velocity (a= 0 )
N = mg
👉case - 3
when lift is moving upward with uniform acceleration
N - mg = ma
N = ma+mg
N = m(a+g )
👉case-4
when lift is moving downward with uniform acceleration
mg - N = ma
N= mg - ma
N = m(g-a)
case -5
if control cable is brakes
a= g
N = m(g-g) It act as a free falling body
N= 0
case- 6 
If downward Acceleration
a>g
g<a
mg<ma
mg-ma<0
m(g-a) <0
N<0
pulley related problem
note -: if pulley is mass less , net force on it is 0 even if it is accelerated
question -: 1
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in figure . what is the action on the floor (normal reaction) by the man in two cases ? if the floor yields to a normal force of 700 N , which mode should the man adopt to lift the block without the floor yielding ?
soln -: In mode (a) the man applies a force equal to 25 kg weight in upward direction A/c newton 's third law there will be downward force of reaction on the floor (normal reaction force on the floor )
Total action on the floor (normal reaction) 
50g + 25g = 75 g = 75*9.8 N = 735 N Ans.
In mode (b) The man applies a downward force equal to 25 kg .wt . according to newton ' third law of motion the reaction will
be in upward direction .
Total reaction (normal reaction) on the floor by the man
50kg.wt - 25 kg.wt = 25 kg.wt = 25*9.8 N = 245 N Ans.
Find the acceleration of both block ?
free body diagram for pulley.
Fnet = ma
2T2-T1 = 0*a
2T2= T1 -------- equ (1)
free body diagram for the block 2M
Fnet = ma
2Mg - T1 = 2Ma
2Mg -2T2 = 2Ma
Mg - T2 = Ma --------- eqn (2) free body diagram for the bock of mass M along incline
For perpendicular motionFnet = 0 N- Mgsin30 degree = 0 N = Mgsin30 degree
2Mg -2T2 = 2Ma
Mg - T2 = Ma --------- eqn (2)
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