\(\vec{F} \ and \ \vec{v}\)are perpendicular
So, Power \(=\vec{F} \cdot \vec{v}=F v \cos 90^{\circ}\)
= Zero  (∵ cos 90° = 0)

he downward force on the elevator is
F = m g + F _f = (1800 x 10) + 4000 = 22000 N
The motor must supply enough power to balance this force. Hence,
P = F. v = 22000 x 2 = 44000 W
= 59 hp

Here, F = \((-\hat{i}+\hat{2j}+\hat{3k})\) N and s =  \((\hat{4k})\) m
W = F.s = F_z.S_z 
= 3 \(\times\) 4
= 12N - m
= 12 J

Loss in kinetic energy on collision is  \(\frac{1}{2}\left(\frac{m_1m_2}{m_1+m_2} \right)\mu^2\)

(i) The part correctly describes the trajectory of the stone after the spring breaks because when a stone tied to one end of a string is whirled round in a circle then velocity of the stone at any point is along the tangent at that point. If the string breaks suddenly, then stone flies off tangentially, along the direction of its velocity.
(ii) The part correctly describes the trajectory of the stone after the spring breaks because when a stone tied to one end of a string is whirled round in a circle then velocity of the stone at any point is along the tangent at that point. If the string breaks suddenly, then stone flies off tangentially, along the direction of its velocity.
(iii) The part correctly describes the trajectory of the stone after the spring breaks because when a stone tied to one end of a string is whirled round in a circle then velocity of the stone at any point is along the tangent at that point. If the string breaks suddenly, then stone flies off tangentially, along the direction of its velocity.

No change in speed, but change in direction is possible. Forces acting on a body in circular motion is an example.

Let a stone of mass m be dropped from a point A at a height h.
P.E. at A = mgh
K.E. = 0
Total energy at A = mgh
As it reaches B, it would have lost some P.E. and gained K.E.
Velocity on reaching

\(\mathrm{B}=\sqrt{2 g x}\)
P.E. at B = mg (h - x)
\(\mathrm{K.E.}=\frac{1}{2} m v_{B}^{2}=\frac{1}{2} m \cdot 2 g x=m g x\)
Total energy at
B = mg (h - x) + mgx = mgh
On reaching the ground C the mass must have gained a velocity \(\sqrt{2 g x}\) and the P.E must be zero.
P.E. at C = 0
\(\text { K.E. at } \mathrm{C}=\frac{1}{2} m v_{C}^{2}=\frac{1}{2} m(2 g h)=m g h\)
Total energy at C = mgh
Thus it is proved that the total energy at any point in its path is mgh.

Work done on the cycle by the road is the work done by the stopping (frictional) force on the cycle due to the road.
(a) The stopping force and the displacement make an angle of 180^o (π rad) with each other
Thus, work done by the road,
W_r = Fd cosθ
= 200 x 10 x cos π
= – 2000 J
It is this negative work that brings the cycle to a halt in accordance with WE theorem.
(b) From Newton’s Third Law an equal and opposite force acts on the road due to the cycle. Its magnitude is 200 N. However, the road undergoes no displacement. Thus, work done by cycle on the road is zero.

Initial momentum of each ball before collision.
= 0.05 x 6 kg m s^-l = 0.3 kg m s^-l
Final momentum of each ball after collision
= - 0.05 x 6 kg m s^-1 = - 0.3 kg ms^-1
Impulse imparted to each ball due to the other
=final momentum - initial momentum
= - 0.3 kg ms^-1 - 0.3 kg ms^-1
=- 0.6 kg ms^-1
= 0.6 kg ms^-1 (in magnitude)
The two impulses are opposite in direction.

(a) Potential energy of the body decreases, because the body in this case goes closer to the centre of the force.
(b) Kinetic energy, because friction does its work against motion.
(c) Internal forces can not change the total or net momentum of a system. Hence the rate of change of total momentum of many particle system is proportional to the external force on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy / total linear momentum/ total energy of the system of two bodies. 

   
Initial mass of the rocket, m = 20000 kg
Initial acceleration a = 5.0m/s²  in upwards direction
Let initial thrust of the blast be T.
T - mg = ma
or T = mg + ma
= m(g + a)
= 20000 x (9.8+ 5.0)
= 2 x 10⁴ x 14.8N
= 29.6 x 104 N = 2.96 x 10⁵ N

Mass of stone, m = 0.25 kg, Radius of the string, r = 1.5 m
Frequency, v = 40rev/min = \(\frac { 40 }{ 60 } \) rev/s = \(\frac { 2 }{ 3 } \) rev/s
Centripetal force required for circular motion is obtained from the tension in the string.
\(\therefore \) Tension in the string = Centripetal force
T = \(mr{ \omega }^{ 2 }\)
\(= mr\left( 2\pi n \right) ^{ 2 } \ \left[ \therefore \omega =2\pi v \right] \)
\(= mr4\pi ^{ 2 }{ v }^{ 2 }\)
\(T=0.25\times 1.5\times 4\times \left( \frac { 22 }{ 7 } \right) ^{ 2 }\times \left( \frac { 2 }{ 3 } \right) ^{ 2 }= 6.6N\)
Maximum tension which can be withstand by the string
\({ T }_{ max }=200\quad N=\frac { mv^{ 2 }max }{ r } \)
\(\\ { v }_{ max }=\sqrt { \frac { { T }_{ max }\times r }{ m } } =\sqrt { \frac { 200\times 1.5 }{ 0.25 } } =34.6{ m }/{ s }\)

Suppose the point O as the position of bat. AO line shows the path along which the ball strikes the bat with velocity u and OB is the path showing deflection such that \(\angle\)AOB
= 45°. Now initial momentum of ball 
= mu cos \(\theta\)
\(={0.15\times 54\times 1000\times cos 22.5\over 3600}\)
= 0.15 x 15 x 0.9239 along NO
Final momentum of ball = mu cos\(\theta\) along ON

Impulse = change in momentum = mu cos \(\theta\) - (- mu cos \(\theta\))
= 2 mu cos \(\theta\) =2 x 0.15 x 15 x 0.9239 = 4.16 kg m^-l.

Given: The mass of the body is 5 kgand the two perpendicular forces are 8 N and 6 N.
The forces on the body are shown below:

The resultant force is given as,
F R = 8 2 + 6 2 = 64+36 =10 N
The tangent of the angle is given as,
tanθ= 6 8 θ= tan ⁻¹ ( 3 4 ) =37°
Newton’s second law of the motion is given as,
F=ma
By substituting the given values in the above expression, we get
10 N=( 5 kg )a a=2 m/ s 2
Thus, the magnitude of acceleration is 2 m/ s 2 and its direction is along the direction of the resultant force. The direction of resultant force is at an angle of 37° with 8 N force.

(a) We know that \({ \mu }_{ k }=\frac { frictional\ force }{ normal\ reaction } \)
\(\therefore\) frictional force = \({ \mu }_{ k }\times normal\ reaction\)
= 0.1 \(\times\) 2 kg wt = 0.1 \(\times\) 2 \(\times\) 9.8 N = 1.96 N
net effective force = (7-1.96)N = 5.04 N
acceleration = \(\frac { 5.04 }{ 2 } { ms }^{ -2 }=2.52\ { ms }^{ -2 }\)
distance, \(s=\frac { 1 }{ 2 } \times 2.52\times 10\times 10=126\ m\)
work done by applied force = 7 \(\times\) 126 J = 882 J
(b) Work done by friction = 1.96 \(\times\) 126 = -246.96 J
(c) Work done by net force = 5.04 \(\times\) 126 = 635.04 J
(d) Change in the kinetic energy of the body
= work done by the net force in 10 seconds
= 635.04 J (This is in accordance with work-energy theorem).

(a) Positive In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket.
(b) Negative In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative.
(c) Negative Since the direction of frictional force is opposite to the direction of motion, the work done by frictional force is negative in this case.
(d) Positive Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.
(e) Negative The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, the work done is negative in this case.

(a) False, in elastic collision the linear momentum and kinetic energy of the system as whole is conserved, but momentum and kinetic energy of each of individual body change.
(b) False, Although internal forces are balanced, they cause no work to be done on a body. It is the external forces that have the ability to do work. Hence, external forces are able to change the energy of a system.
(c) False, The work done in the motion of a body over a closed loop is zero for a conservation force only
(d) It is usually true but not always true.As an exmple in the exlposion of a cracker final kinetic energy is greater than the initial kinetic energy. Again final kinetic energy of gun-bullet system after firing is more than initial kinetic energy before collision.

When any object is thrown from a train, the influence of train on the object becomes zero at the same moment, i.e. there is no effect of acceleration of train on the object.
\(\because \)  Mass of  stone (m) = 0.1 kg
(i) When stone is dropped from the window of a stationary train, it falls freely under gravity.
\(\therefore \)  Net force acting on stone (F)
= mg = 0.1 x 10
= 1.0 N (vertical downward)
(ii)Just after the stone is dropped from the window of a train running at a constant velocity, i.e. acceleration of the train is zero. So, no force acts on the stone due to motion it falls freely under gravity.
The force acts on it is due to its weight only,
i.e. acceleration of the train is zero, so no force acts, on the stone due to this motion.
(iii)The train accelerate horizontally by 1 m/s², but as the stone is left it moves under gravity only so the net force on it will be due to gravity only.
(iv) Lying on the floor of a train which is accelerating with 1 m/s² , the stone being at rest relative to the train. Neglect air resistance throughout.

Force F = ma, therefore force acting on a particle in unaccelerated (a = 0) motion is zero.
(i) As drop of rain is falling downward with a constant speed, therefore its acceleration is zero.
According to Newton's second law of motion, net force acting on drop F = ma = 0.
(ii) In floating condition, the weight of the body is balanced by the upthrust. Therefore, net force acting on a cork floating on water = 0
(iii) As kite is held stationary in the sky, therefore acceleration of the kite is zero. Therefore, net force acting on the car F = ma = 0.
(iv) As car is moving with a constant velocity, therefore, its acceleration is zero. i.e., a = 0 therefore, net force acting on the car F = ma = 0.
(v) As electron is in a space where there is no electric field, magnetic field and gravitational (material) objects, therefore, no electric, magnetic and gravitational force is acting on it. Hence, net force acting on electron is zero.

We know, y = ut + \(\frac { 1 }{ 2 } \) gt²
Now, v = \(\frac { dy }{ dt } \) = u + gt
acceleration, a = \(\frac { dv }{ dt } \) = g
Then, the force is given by
F = ma = mg
Thus, the given equation describes the motion of a particle under acceleration due to gravity and y is the position coordinate in the direction of g.

(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.

(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
On a rainy day, the roads are wet. Wetting of roads lowers the coefficient of friction between the types and the road. Therefore, grip on a road of car reduces and thus chances of skidding increases.