40 N

Since \(tan \theta =\mu =\sqrt{3}\)
\(\theta =tan ^{-1}(\sqrt{3})\Rightarrow \theta =60^o\)

When the monkey is falling freely, it would be a state of weightlessness. So, tension will be zero and the rope will not break.

When the monkey climbs up with uniform speed, then
T = mg = 40 kg x 10 ms^-2 = 400 N
The rope will not break.

When the monkey is climbing down with an acceleration, then

mg - T=ma
\(\Rightarrow\)T=mg - ma = m (g - a)
or T=40 kg x (10 - 4) ms^-2 = 240 N
The rope will not break.

When the monkey climbs up with an acceleration a, then
T - mg = ma
where T represents the tension
\(\therefore T=mg+ma=m(g+a)\)
or T = 40 kg (10 + 6) ms^-2 = 640 N
But the rope can withstand a maximum tension of 600 N. So the rope will break.

\(1/2\sqrt { 3 }\) 

Given, m = 2 kg, u = 2ms^-1,\(\mu \) = 0.2
Force of friction, F  \(= \mu R\)
\( F=\ \mu mg\ \ \left[ \because R\ =mg \right] \)
\(F=\quad 0.2\times 2\times 9.8\)
\( F=\ 3.92Nx = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Distance moved per second s = ut
s = \(2\times 1=2\)
Work done by friction per second, W = Fs
W = 3.92\(\times\)2 = 7.84 J
Heat produced,    \(H=\frac { W }{ J } \Rightarrow \quad H=\frac { 7.84 }{ 4.2 } \)
H = 1.87 cal

Here, radius r =20 cm
Centripetal acceleration, = 980 cms^-2
We know that centripetal acceleration, a = rω²
\(\omega =\sqrt { \frac { a }{ r } } =\sqrt { \frac { 980 }{ 20 } }\)
\(\omega =\sqrt { 49 } =7rad/s\)

Given, velocity, v = 50 ms^-1
Radius, r = 250 m
Centripetal accleration, a = v²/r
a = \(\frac{50 \times 50}{250}\)
= 10 ms^-1

As the applied force increases, the static friction also increases and becomes equal to the applied force to make the object stationery. That is why static friction is called a self-adjusting force.

25.6 m/s

Given, mass, m = 72.2 kg
Gravity acceleration, g = 9.8 m/s² 
Scale reading = apparent weight = R = ?
(i)  While descending with constant velocity, a = 0
\(\therefore \)    R =  mg
R = 72.2 x 9.8
\(\Longrightarrow \)  R = 707.56 N
(ii)  While ascending with a = 3.2 m/s² 
R = m ( g + a )
R = 72.2 ( 9.8 + 3.2 ) = 938.6 N

Given, F = 36 dyne at an angle of 6000 .
∴∴   Component of force along x-direction
F_x = F cos 6000 = 36 x 1/2 = 18 dyne
 But  F_x = ma_x ,
a_x = \(\frac{F_x}{m} =\frac {18}{18} = 1\) cm/s²

Since the point P is at rest.

\(F_{2}=1 N \sin 45^{\circ}+2 N \cos 45^{\circ}=\frac{3}{\sqrt{2}} N\)
As F₁ + 1N cos 45° = 2Nsin 45^o
\(F_{1}=\frac{2}{\sqrt{2}} N-\frac{1}{\sqrt{2}} N=\frac{1}{\sqrt{2}} N\) 

The acceleration produced in the body of mass 10 + 20 = 30 kg is given by,
\(a=\frac{F}{m}=\frac{500}{30}=\frac{50}{3} \mathrm{~ms}^{-2}\)
When 500 N pull is applied on 20 kg, tension, TI produced is given by,
T₁ = m₁a
\(=10 \times \frac{50}{3}=\frac{500}{3}=166.7 \mathrm{~N}\)
When 500 N pull is applied on 10 kg, tension T2 produced is given by,
T₂ = m₂a
\(=20 \times \frac{50}{3}=\frac{1000}{3}=333.4 \mathrm{~N}\)
Thus the tension depends mass-end on which the pull is applied.

(i) The net force f₁ exerted on the trailer in the direction of f₁ = (T - 1000)N, where T is tension in the connecting rope.
ஃ T - 1000 = 2000 a           ..............(i)
Similarly, the net force f₂ exerted by the engine of the truck is given by
f₂ = (6000 - 500 - T) = 1000 a
or 5500 - T = 1000 a           ..............(ii)
Adding (i) and (ii), we have,
4500 = 3000 a
or \(a=\frac{4500}{3000}=1.5 \mathrm{~ms}^{-2}\)
(ii) Putting the value of 'a' in (i) we have
T = 1000 + 2000 a
= 1000 + 2000 x 1.5
= 4000 N

0.6

(i) Let a be the acceleration of the system to right. All the three frictional forces f₁=μm₁g, f₂ = μm₂g, and f₃ =μm₃g will be directed to the left as the motion of bodies is to the right. Hence, for the whole system

F-μm₁g-μm₂g-μm₃g=(m₁+m₂+m₃)a
F= (m₁+m₂+m₃)(a+μg)
=(1.5 + 2 + 1)(3 + 0.2 x 9.8) = 22.3 N
(ii) The force exerted by the 1.5 kg block on the 2 kg block = F - m₁ (a + μg)
= 22.3 - 1.5 (3 + 0.2 x 9.8)
= 22.3 - 7.44 = 14.86 N

Considering components of tensions T₁ and T₂ along the Horizontal and vertical directions,
we have
- T₁cos \(\alpha\) + T₂ cos \(\beta\) =0 or
T₁ cos \(\alpha\) T₂ cos \(\beta\)............ (i)
and T₁ sin \(\alpha\) + T₂ sin \(\beta\) mg ............(ii)
From (i) T₂ =\({T_1cos \alpha \over cos \beta}\) and substituting it in(ii), we get
\(T_1 sin \alpha +({T_1cos \alpha \over cos \beta})sin \beta =mg \ or \ t_1[{ sin \alpha cos \beta +cos \alpha sin \beta \over cos \beta}]=mg\)
\(and \ hence \ T_2 ={T_1 cos \alpha \over cos \beta }={mg \ cos \beta \over sin (\alpha +\beta)}.{cos \alpha\over cos \beta}={mg \ cos \alpha \over sin (\alpha +\beta)}\)

Here, \(m=10 \mathrm{~kg}, \theta=30^{\circ}, \mu=\frac{1}{\sqrt{3}}\)
As is clear from force required just to push the body up the inclined plane is
\( F=m g \sin \theta+f \)
\(=m g \sin \theta+\mu R \)
\(=m g \sin \theta+\mu m g \cos \theta \)
\(=m g(\sin \theta+\mu \cos \theta) \)
\(=10 \times 9.8\left(\sin 30^{\circ}+\frac{1}{\sqrt{3}} \cos 30^{\circ}\right) \)
\( F=98\left(0.5+\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2}\right)=98 N \)

\(\because \)  Mass of each coin = m
Number of total coins = 10
(iii) Reaction of the 6th coin on the 7th coin
= -(force exerted on 6th coin)
= -(weight of 4 coins)
= -4 mg N (vertically upward)

\(\because \)  Mass of each coin = m
Number of total coins = 10
Force acting on 7th coin by the 8th coin = weight of the 8 coins + weight of two coins supported by  8 coins.
= mg + 2 mg
= 3 mg N ( downward)

\(\because \)  Mass of each coin = m
Number of total coins = 10
(i) Force acting on 7th coin (counted from the bottom)
= Weight of the coins above it
= Weight of 3 coins
= 3 mg N (downward)
(ii) Force acting on 7th coin by the 8th coin = weight of the 8 coins + weight of two coins supported by  8 coins.
= mg + 2 mg
= 3 mg N ( downward)
(iii) Reaction of the 6th coin on the 7th coin
= -(force exerted on 6th coin)
= -(weight of 4 coins)
= -4 mg N (vertically upward)

When an object is thrown vertically upwards or it falls vertically downward under gravity, then an acceleration  g = 10 m/s² acts downward due to the earth's gravitational pull.
Mass of pebble (m) = 0.05 kg
(iii) At the highest point
Net force acting on pebble
(F) = ma = 0.05 x 10 N
= 0.50 N  (vertically downward)
If  pebble was thrown at an angle of 45\(^0\)   with the horizontal direction, then acceleration acting on it and therefore force acting on it wil remain unchanged, i.e. 0.50 N (vertically downward). In case (c), at the highest point the vertical component of velocity will be zero but horizontal component of velocity will not be zero.

Radius of circular bend, r = 30m
Speed of the train, v = 54 km/h
=  \(54\times \frac { 5 }{ 18 } { m }/{ s\ \quad \left[ \because 1km/h=\frac { 5 }{ 8 } { m }/{ s } \right] }\)
= 15 m/s
Let \(\theta\) be the angle of banking required to prevent wearing out the rails, then
 \(\tan { \theta =\frac { { v }^{ 2 } }{ rg } } =\frac { (15{ ) }^{ 2 } }{ 30\times 9.8 } \)
\(=\quad \frac { 225 }{ 30\times 9.8 } =0.7653\)
\(\\ \theta =\tan { ^{ -1 }\quad (0.7653)=37.{ 4 }^{ o } }\) 

\(\because \)  Mass of the helicopter, m₁ = 1000 kg
Mass of the crew and the passengers, m₂ = 300 kg
Acceleration of the helicopter, a = 15 m/ s²
Acceleration  due to gravity, g = 10 m/s²
(i) Let R, be the reaction applied by the floor on the crew and the passengers.

R₁- m₂g = m₂a
or
R₁= m₂g + m₂a = m₂(g+a)
= 300(10 + 15) = 7500N(upward direction)
(ii) Action of the rotor of the helicopter on the surrounding air
= ( m₁ + m₂ ) g + ( m₁ + m₂ ) a
= ( m₁ + m₂ ) ( g + a )
= ( 1000 + 300 ) x (10 + 15 )
= 1300 x 25 = 32500 N
Force (action ) of the rotor of the helicopter on the surrounding air = 32500 N (downward)

(i) 5.2 N backward
(ii) 450 N on the boat
(iii) 180 N on the hockey stick