When the lift is moving down with an acceleration
' a ' the weight of the person W is, given by
W = mg = ma = m(g-a)
Fractional change is the weight of the person
= \(\frac { mg-w }{ mg } \)
= \(1-\frac { w }{ mg } -\frac { mg-(mg-ma) }{ mg } \)

 \(=\frac { a }{ g } =\frac { 5.0 }{ 9.8 } =0.51\)
=Percentage change = 0.51\(\times\)100 
= 51 % 

Force F = F₁ = F₂
F₁ = \(\frac { dp }{ { dt }_{ 1 } } ,{ F }_{ 2 }=\frac { dp }{ { dt }_{ 2 } } \)
\({ F }_{ 2 }={ F }_{ 1 }=\frac { { dt }_{ 1 } }{ { dt }_{ 2 } } \)
\(=10\times \frac { 8 }{ 4 } =20N\)

When a track goes round a level curve, the force of friction between the tyres and the road provides the necessary centripetal force. If the frictional force, which acts as centripetal force and keeps the body moving along the circular road is not enough to provide the necessary centripetal force, the car will skid. In order to avoid skidding while going round a curved path the outer edge of the road is raised above the level of the inner edge. This is known as banking of curved roads or tracks.
Let the surface of the road makes angle ፀ with horizontal surface. Then the normal force makes the same angle d with the vertical. When the car takes a turn, there are two forces acting on the car:
(a) Gravitational force mg (downwards)
(b) Normal force N (perpendicular to surface).
We can resolve the normal force into two components. N cos ፀ and N sin ፀ as shown in Figure. The component N cos ፀ balances the downward gravitational force 'mg' and component N sin ፀ will, provide the necessary centripetal acceleration. By using Newton second law
N cos ፀ = mg .....(1)
N sin ፀ = \(\frac { { mv }^{ 2 } }{ r } \)  ....(2)
Eq (2) / (1) ⇒ tan ፀ = \(\frac { { v }^{ 2 } }{ rg } \)
v= \(\sqrt { rg\tan\theta } \)
Need Banking of Tracks
1. The banking angle ፀ and radius of curvature of the road or track determines the safe speed of the car at the turning.
2. If the speed of car exceeds this safe speed, then it starts to skid outward but frictional force comes into effect and provides an additional centripetal force to prevent the outward skidding.
3. At the same time, if the speed of the car is little lesser than safe speed, it starts to skid inward and frictional force comes into effect, which reduces centripetal force to prevent inward skidding. 
4. However if the speed of the vehicle is sufficiently greater than the correct speed, then frictional force cannot stop the car from skidding.

Static friction: The opposing force that comes into play when one body tends to move over the surface of another, but the actual motion has yet not started is called static friction.
Limiting friction: If the applied force is increased the force of static friction also increases. If the applied force exceeds a certain (maximum) value, the body starts moving. This maximum value of static friction up to which body does not move is called limiting friction.
Kinetic or dynamic friction: If the applied force is increased further and sets the body in motion, the friction opposing the motion is called kinetic friction.
We can reduce friction
(1) By polishing.
(2) By lubrication.
(3) By proper selection of material.
(4) By streamlining the shape of the body.
(5) By using ball bearing.

(i) The car is moving with uniform velocity v with respect to a person standing (at rest) on the ground.
(ii) As the car is moving with constant velocity with respect to ground to the person is at rest on the ground, both frames (with respect to the car and to the ground) are inertial frames.

(i) Inertia of rest: When a stationary bus starts to move, the passengers experience a sudden backward push.
(ii) Due to inertia, the body (of a passenger) will try to continue in the state of rest, while the bus moves forward. This appears as a backward push.

Yes, μ > 1, friction is stronger than normal force

The surface area of parachute is very large. And when it descends downwards, the air provides resistance to it and so it descends slowly.

(i) Due to excessive polishing, more molecules of the two surfaces come closer. Thus they form more bonds and offer great resistances.
(ii) Because friction arises due to molecular adhesion i.e. due to electrostatic forces.

(a) A frame of reference which is at rest or which is moving with a uniform velocity along a straight line is called an inertial frame of reference.
(b) In the inertial frame of reference Newton's laws of motion holds good.
Example: The lift at rest, lift moving (up or down) with constant velocity, car moving with constant velocity on a straight road.

The centrifugal force appears to act on the particle, only when we analyses the motion from a rotating frame. With respect to an inertial frame there is only centripetal force which is given by the tension in the string. For this reason centrifugal force is called as a 'pseudo force'. A pseudo force has no origin, It arises due to the non-inertial nature of the frame considered.

(i) Consider the firing of a gun. Here the system is Gun+bullet. Initially, the gun and bullet are at rest, hence the total linear momentum of the system is zero. Let \(\vec { { p }_{ 1 } } \) be the momentum of the bullet and \(\vec { { p }_{ 2 } } \) the momentum of the gun before firing. Since initially both are at rest, \(\vec { { p }_{ 1 } } \)=0 , \(\vec { { p }_{ 2 } } \)=0
(ii) Total momentum before firing the gun is zero,\(\vec { { p }_{ 1 } } +\vec { { p }_{ 2 } } =0\)
(iii) According to the law of conservation of linear momentum, total linear momentum has to be zero after the firing also.
(iv) When the gun is fired, a force is exerted by the gun on the bullet in forward direction. Now the momentum of the bullet changes from \(\vec { { p }_{ 1 } } \) to \(\vec { { p }_{ 1 } } \)  To conserve the total linear momentum of the system, the momentum of the gun must also change from \(\vec { { p }_{ 2 } } \) to \(\vec { { p }_{ 2 } } \) Due to the conservation of linear momentum \(\vec { { p }'_{ 1 } } +\vec { { p }'_{ 2 } } =0\) It implies that  \(\vec { { p }'_{ 1 } } =-\vec { { p }'_{ 2 } } \) the momentum of the gun is exactly equal, but in the opposite direction to the momentum of the bullet.
(v) This is the reason after firing, the gun suddenly moves backward with the momentum \(\left( \vec { { p }'_{ 2 } } \right) \) It is called 'recoil momentum'. This is an example of conservation of total linear momentum.

(i) When a vehicle travels in a curved path, there must be a centripetal force acting on it.
(ii) This centripetal force is provided by the frictional force between tyre and surface of the road.
(iii) Consider a vehicle of mass Om'moving at a speed 'v' in the circular track of radius or
(iv) There are three forces acting on the vehicle when it moves.
(1) Gravitational force (mg) acting downwards
(2) Normal force (mg) acting upwards
(3) Frictional force (F_s) acting horizontally Inwards along the road.
(v) Suppose the road is horizontal then the normal force and gravitational force are exactly equal and opposite.
(vi) The centripetal force is provided by the force of static friction F between the tyre and surface of the road Which acts towards the center of the circular track.
\(\frac { { mv }^{ 2 } }{ r } ={ F }_{ s }\quad \)
The static friction can increase from zero to a maximum value
\(\quad { F }_{ s }\le { U }_{ s }mg\)
(vii) The static friction would be able to provide necessary centripetal force to bend the car on the road. So the coefficient of static friction between the tyre and the surface of the road determines what maximum speed the car can have for safe turn.
\(if\frac { { mv }^{ 2 } }{ r } >{ u }_{ s }mg\quad or\quad { u }_{ s }<\frac { { V }^{ 2 } }{ rg } \quad \)
If the static friction is not able to provide enough centripetal force to turn, the vehicle will start to skid

(i) Consider the case of a whirling motion of a stone tied to a string. Assume that the stone has angular velocity ω in the inertial frame (at rest).
(ii) If the motion of the stone is observed from a frame which is also rotating along with the stone with same angular velocity ω then, the stone appears to be at rest.
(iii) This implies that in addition to the inward centripetal force - mω²r there must be an equal and opposite force that acts on the stone outward with value + mω²r.
(iv) So the total force acting on the stone in a rotating frame is equal to zero (-mω²r + mω²r=0).
(v) This outward force + mω²r is called the centrifugal force.