Mass m = 5 kg
K.E E = 1000 J
g = 10 ms^-2
At a height 'h', mgh = \(\frac { E }{ 2 } \)
\(5\times 10\times h=\frac { 1000 }{ 2 } \)
\(h=\frac { 500 }{ 50 } =10\ m\)

a=\(\frac { { m }_{ 1 }-{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \times g=\frac { 5-4 }{ 5+4 } \times 10=\frac { 1 }{ 9 } \)=1.1 ms^-2.

(i) Displacement is zero or it is perpendicular to force.
(ii) Conservative force moves a body over a closed path.

To reduce friction and so to reduce wear and tear.

The unit of power is watt. One watt is defined as the power when one joule of work is done in one second.

As F = rate change of momentum
F = 1 kg-m/s² = 1N

Rolling friction is less than sliding friction.

When the object is at rest, the external force and the static frictional force are equal and opposite.
The magnitudes of these two forces are equal f_s=F_ext
Therefore, the static frictional force acting on the object is
f_s = 5N
The direction of this frictional force is opposite to the direction of F_ext.

We use Newton's second law, \(\vec { F_{ net } } =m\vec { a } \) where \({ \vec { F } }_{ net }=\vec { { F }_{ 1 } } +\vec { { F }_{ 2 } } \). From the above equations the acceleration is \(\vec { a } =\frac { \vec { { F }_{ net } } }{ m } \), where
\(\vec { { F }_{ net } } =(5+3)\hat { i } +(8-4)\hat { j } +(7+3)\hat { k } \)
\({ \vec { F } }_{ net }=8\hat { i } +4\hat { j } +10\hat { k } \)
\(\vec { a } =\left( \frac { 8 }{ 2 } \right) \hat { i } +\left( \frac { 4 }{ 2 } \right) \hat { j } +\left( \frac { 10 }{ 2 } \right) \hat { k } \)
\(\vec { a } =4\hat { i } +2\hat { j } +5\hat { k } \).

Force \(\vec F=\vec i+2\vec j+3\vec k\)
Distance \(\vec S=4\vec i+6\vec j\)
Work done  \(\vec F.\vec S=(\vec i2\vec j+3\vec k).(4\vec i+6\vec j)=4+12+0=16J\)

Work done, \(W-\int^{x_f}_{x_i}F(x)dx=k\int_0^4x^2 dx={64\over 3}Nm\)

Force is a vector and magnitude of the vector is represented by the length of the vector. Here \(\vec { { F }_{ 1 } } \) has greater length compared to other two. So \(\vec { { F }_{ 1 } } \) is largest of the three.

T = \(\frac { mv^{ 2 } }{ R } =\frac { 1\times (10)^{ 2 } }{ 1 } \) = 100N

Formula: F=\(V\frac { dm }{ dt } \) = 4\(\times\)3 =12N
∴ \(a=\frac { F }{ m } \) = \(\frac { 12 }{ 5 } \)= 2.4 m/s²

The first body is very much lighter than the second body  \(\left( { m }_{ 1 }<<{ m }_{ 2 },\frac { { m }_{ 1 } }{ { m }_{ 2 } } <<1 \right) \)then the ratio \(\frac { { m }_{ 1 } }{ { m }_{ 2 } } \approx 0\) and also if the target is at rest (u₂=0)
Dividing numerator and denominator of equation \({ v }_{ 1 }=\left( \frac { { m }_{ 1 }-{ m }_{ 2 } }{ { { m }_{ 1 }+m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { 2m }_{ 2 } }{ { { m }_{ 1 }+m }_{ 2 } } \right) { u }_{ 2 }\) by m₂ we get
\({ v }_{ 1 }=\left( \frac { \frac { { m }_{ 1 } }{ { m }_{ 2 } } -1 }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) { u }_{ 1 }+\left( \frac { 2 }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) \left( 0 \right) \)
\({ v }_{ 1 }=\left( \frac { 0-1 }{ 0+1 } \right) { u }_{ 1 }\)
v₁=-u₁
Similarly, dividing numerator and denominator of equation 
\({ v }_{ 2 }=\left( \frac { { 2m }_{ 1 } }{ { { m }_{ 1 }+m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { m }_{ 2 }-{ m }_{ 1 } }{ { { m }_{ 1 }+m }_{ 2 } } \right) { u }_{ 2 }\), by m_2,  we get
\({ v }_{ 2 }=\left( \frac { 2\frac { { m }_{ 1 } }{ { m }_{ 2 } } }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) { u }_{ 1 }+\left( \frac { 1-\frac { { m }_{ 1 } }{ { m }_{ 2 } } }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) \left( 0 \right) \)
\({ v }_{ 2 }=\left( 0 \right) { u }_{ 1 }+\left( \frac { 1-\frac { { m }_{ 1 } }{ { m }_{ 2 } } }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) \left( 0 \right) \)
v₂=0 

A force is said to be non-conservative if the work done by or against the force in moving a body depends upon the path between the initial and final positions. This means that the value of work done is different in different paths.

10⁷ erg = 1 Joule
1 erg = 10^-7 Joule

Work is said to be done by the force when the force applied on a body displaces it. Eg: Horse pulls a cart, engine pulls a train.

When a rubber ball hits a massive object, say, earth, the ball is distorted. A large amount of heat is generated in the ball by the rubbing of the rubber molecules against each other. This effect is essentially absent in a hard material. So, a metal ball would often lose less energy upon collision than would a rubber ball.

The work done in striking the two stones against each other gets converted into heat. This appears as a spark.

Here, F =20 kg wt = 20\(\times\)9.8 N,
S = 20m, \(\theta\) = 60°
W = F, \(\cos\theta\) = 20\(\times\)9.8\(\times\)20\(\times\)cos60°
=20\(\times\)9.8\(\times\)20\(\times\)0.5 = 1960J

m = 60 + 20 = 80 kg
h = 20\(\times\)0.2 = 4m
g = 9.8 ms^-2, t = 10s
\(P={W\over t}={mgh \over t}={80\times 9.8 \times 4 \over 10}={3136 \over 10}=313.6W\)

P.E of the spring when stretched through a distance x,
\(U={1\over2}{kx}^{2}=10J\)
When x becomes lx, the potential energy will be
\(u'={1\over2}k(2x)^2=4\times{1\over2}{kx}^{2}=4\times10=40J\)
\(\therefore\) Workdone = u' - u = 40 - 10 = 30 J

Action and reaction acts on different bodies.

Force of friction always opposes the relative motion. ∴ θ = 180⁰

It is the force of reaction on the body due to the surface on which it is placed. It always acts perpendicular to the point of contact.

Frictional force which always opposes the relative motion between an object and the surface where it is placed.

(i) Gravitational force (mg) acting downwards
(ii) Normal force (mg) acting upwards
(iii) Frictional force (F_s) acting horizontally Inwards along the road.

Downward gravitational force (mg), Normal force perpendicular to inclined surface (N).

(i) Second biker cannot apply Newton's second law, because he is moving with acceleration \(\overset { \rightarrow }{ a } \) with respect to Earth (he is not in inertial frame).
(ii) But the first biker can apply Newton's second law because he is moving at constant velocity with respect to Earth (he is in inertial frame).

Free body diagram is a simple tool to analyse the motion of the object using Newton's laws.
The following systematic steps are followed for developing the free body diagram:
(i) Identify the forces acting on the object.
(ii) Represent the object as a point.
(iii) Draw the vectors representing the forces acting on the object.

The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.

The necessary condition is \(\frac{mv{2}}{r} \ge mg\) i.e. v\(\ge \sqrt{rg}\) at the highest point and \(v>\sqrt{5gr}\) at the lowest point.

The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from ground to that height h with constant velocity. 

(i) Potential energy of an object at a point P is defined as the amount of work done by an external force in moving the object at constant velocity from the point 0 (initial location) to the point P (final location).
(ii) At initial point O potential energy can be taken as zero.
(iii) Mathematically, potential energy is defined as \(U=\int { { \vec { F } }_{ a } } .\vec { dr } \)
where the limit of integration ranges from initial, location point O to final location point P.

(i) For an isolated system, the sum of all forms of energy i.e., the total energy remains the same in any process irrespective of whatever internal changes may take place.
(ii) This means that the energy disappearing in one form reappears in another form. This is known as the law of conservation of energy.

During the jump, our feet at once come to rest and for this smaller time (F = Impulse / time) a large force acts on feet. If we bend the knees slowly, the value of time of impact increases and less force acts on our feet. So we get less hurt.

The statement is wrong. The earth revolves around the sun under the force of attraction of the sun. This force (centripetal) is always perpendicular to the motion of the earth. Therefore, \(\theta\) = 90° and W = FS cos 90° = 0.Hence, sun does to work on the earth.

Yes, there is ail internal energy in a body due to the thermal agitation of the particles of the body, while the vector sum of the momenta of the moving particles may be zero.

Every object continues to be in the state of rest or of uniform motion (constant velocity) unless there is external force acting on it.