\(F=-\frac{d u}{d x} \quad F(x) =k x+a x^{3} \)
\(d u =-F d x \)
\(u(x) =-\int_{0}^{x}\left(-k x+a x^{3}\right) d x \)
\(=\int_{0}^{x} k x d x-a \int_{0}^{x} x^{3} d x \)
\(=\frac{k x^{2}}{2}-\frac{a x^{4}}{2} \)
\(U(x) =\frac{x^{2}}{2}\left(k-\frac{a x^{2}}{2}\right) \)
\(u(x)=0 \text { at } x=0 \text { and }\)
\(U(x) =0 ; k-\frac{a x^{2}}{2}=0 \)
\(=\frac{a}{2} x^{2}=-k \)
\(x^{2} =\frac{2 k}{a} \)
\(\therefore x =\sqrt{\frac{2 k}{a}} \)
\(\text { Clearly } u(x)=0 \text { at } x=0 \text { and }\)
\(x=\sqrt{\frac{2 k}{a}}\)
\(\text { For } x>\sqrt{\frac{2 k}{a}} U(x) \text { will be negative. } \)
\(\text { At } x=0 ; F=\frac{-d u}{d x}=0\)
(i.e.,) Slope of V - x graph is zero at x = 0
Hence the most appropriate answer is d.

Radius =R
\(v_{1}^{2}-v_{2}^{2}=4 g R\)
\(\text { Tension } T_{2}=\frac{m v_{2}^{2}}{R_{2}}-m g\)
\(\text {To find minimum speed, let } T_{2}=0\)
\(0 =\frac{m v_{2}^{2}}{R}-m g \)
\(\frac{m v^{2}}{R} =m g \)
\(v_{2}^{2}=R g \ v_{2} =\sqrt{g R} \)
\(\text { sub (2) in the eqn (1) we get }\)
\(v_{1}^{2}-(\sqrt{g R})^{2} =4 g R \)
\(v_{1}^{2}-g R =4 g R \)
\(v_{1}^{2} =4 g R+g R \)
\(=5 g R \)
\(v_{1} =\sqrt{5 g R} \)

For a perfectly elastic collision
\(\left|v_{1}-v_{2}\right|=\left|u_{1}-u_{2}\right| \)
\(v_{e}=\frac{\left|v_{1}-v_{2}\right|}{\left|u_{1}-u_{2}\right|}=1 \)

\(m_{1} =1, \quad m_{2}=2 \)
\(K . E . =m g(h-x) \)
\(\text { For both balls }(h-x)\)
\(=40 \text { i.e.) Same }\)
\(g=\text { constant }\)
\(\therefore K \cdot E_{1}=m_{1} g(h-x)=m_{1} g \times 40 \)
\(K \cdot E_{2}=m_{2} g(h-x)=m_{2} g \times 40 \)
\(\therefore \frac{K \cdot E_{1}}{K \cdot E_{2}}=\frac{m_{1} g \times 40}{m_{2} g \times 40}=\frac{m_{1}}{m_{2}} \)
\( \therefore K \cdot E_{1}: K \cdot E_{2}=1: 2 \)

No, only that much amount of kinetic energy is lost as is necessary for the conservation of momentum.

\(\text {Mass of a bullet } m=20 g=20 \times 10^{-3} \mathrm{~kg} =0.02 \mathrm{~kg}  \)
\(\text {Mass of a pendulum } M =5 \mathrm{~kg}  \)
\(\text {Height } \mathrm{h} =10 \mathrm{~cm}  \)
\(=10 \times 10^{-2}  \)
\(=0.1 \mathrm{~m}  \)

\(\text { K.E. of the block } =\text { P.E. of the block }  \)
\(\frac{1}{2} M v^{2} =M g h  \)
\(\therefore v^{2} =\sqrt{2 g h}  \)
\(=\sqrt{2 \times 9.8 \times 0.1}=\sqrt{1.96}=1.4 \mathrm{~m} / \mathrm{s}  \)
\(\therefore \text { Final speed } v =1.4 \mathrm{~m} / \mathrm{s}  \)
\(\text {Final speed } v =\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}  \)
\(v_{1} =\frac{0.02 u_{1}+5 \times 0}{(0.02+5)}=\frac{0.02}{5.02} u_{1}  \)
\(\text { But } v =1.4  \)
\(\therefore 1.4 =\frac{0.02}{5.02} u_{1}  \)
\(u_{1}=\frac{1.4 \times 5.02}{0.02}=\frac{7.028}{0.02}=351.4 \mathrm{~m} / \mathrm{s}\)
\(\therefore \text { Initial speed } =351.4 \mathrm{~m} / \mathrm{s}\)

Given:
Force mg = 30 N; height = 10 m
Work done to lift a load W = ?
W = F.S (or) mgh
= 30 \(\times\) 10
W = 300J

Power is defined as the rate of work done or energy delivered.
Power(P)=\(\frac{work\ done(W)}{time\ taken(t)}\)

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. 

Let the mass of the first body be m which moves with an initial velocity, u₁ = 10 m s^-1.
Therefore, the mass of second body is 2m and its initial velocity is \({ u }_{ 2 }=\frac { 1 }{ 2 } { u }_{ 1 }=\frac { 1 }{ 2 } (10{ ms }^{ -1 })\)
\({ v }_{ 1 }=\left( \frac { { m }_{ 1 }-{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { 2m }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\)
\({ v }_{ 1 }=\left( \frac { { m }_{ 1 }-2m }{ m+2m } \right) +10+\left( \frac { 2\times 2m }{ m+2m } \right) 5\)
\({ v }_{ 1 }=-\left( \frac { 1 }{ 3 } \right) 10+\left( \frac { 4 }{ 3 } \right) 5=\frac { -10+20 }{ 3 } =\frac { 10 }{ 3 } \)
v₁ = 3.33 ms^-1
\({ v }_{ 2 }=\left( \frac { 2{ m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { m }_{ 2 }-{ m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\)
\({ v }_{ 2 }=\left( \frac { 2m }{ m+2m } \right) 10+\left( \frac { 2m-m }{ m+2m } \right) 5\)
\({ v }_{ 2 }=\left( \frac { 2 }{ 3 } \right) 10+\left( \frac { 1 }{ 3 } \right) 5=\frac { 20+5 }{ 3 } =\frac { 25 }{ 3 } \)
v₂ = 8.33 ms^-1
As the two speeds v₁ and v₂ are positive, they move in the same direction with the velocities 3.33 ms^-1 and 8.33 ms^-1 respectively.

F = k_Ax_A = k_Bx_B
\({ x }_{ A }=\frac { F }{ { k }_{ A } } ,{ x }_{ B }=\frac { F }{ { k }_{ B } } \)
The work done on the springs are stored as potential energy in the springs.
\({ U }_{ A }=\frac { 1 }{ 2 } { k }_{ A }{ x }_{ A }^{ 2 };\quad { U }_{ B }=\frac { 1 }{ 2 } { k }_{ B }{ x }_{ B }^{ 2 }\)
\(\frac { { U }_{ A } }{ { U }_{ B } } =\frac { { k }_{ A }{ x }_{ A }^{ 2 } }{ { k }_{ B }{ x }_{ B }^{ 2 } } =\frac { { { k }_{ A }\left( \frac { F }{ { k }_{ A } } \right) }^{ 2 } }{ { { k }_{ B }\left( \frac { F }{ { k }_{ B } } \right) }^{ 2 } } =\frac { \frac { 1 }{ { k }_{ A } } }{ \frac { 1 }{ { k }_{ B } } } \)
\(\frac { { U }_{ A } }{ { U }_{ B } } =\frac { { k }_{ B } }{ { k }_{ A } } \)
k_A > k_B implies that U_B > U_A.Thus, more work is done on B than A.

(a) The potential energy U = mgh = 2 \(\times\) 10 \(\times\) 5 = 100 J
Here the positive sign implies that the energy is stored on the mass.
(b) This potential energy is transferred from external agency which applies the force on the mass.
(c) The external applied force \(\overrightarrow { { F }_{ a } } \) which takes the object to the height 5 m is \(\overrightarrow { { F }_{ a } } =-\overrightarrow { { F }_{ g } } \)
\(\overrightarrow { { F }_{ a } } =-(-mg\hat { j } )=mg\hat { j } \)
where, \(\hat { j } \) represents unit vector vertical upward direction.
(d) From the definition of potential energy, the object must be moved at constant velocity. So the net force acting on the object is zero.
\(\overrightarrow { { F }_{ g } } +\overrightarrow { { F }_{ a } } =0\)

Relation between power and velocity
The work done by a force \(\overrightarrow{\mathbf{F}}\) for a displacement \(d \vec{r}\) is
\(W=\int \overrightarrow{\mathrm{F}} \cdot d \vec{r}\)  ......(1)
Left hand side of the equation (1) can be written as
\(W=\int d W=\int \frac{d W}{d t} d t\)
(multiplied and divided by dt) (2)
Since, velocity is \(\vec{v}=\frac{d \vec{r}}{d t} ; \overrightarrow{d r}=\vec{v} d t.\) Right hand side of the equation (1) can be written as
\(\int \overrightarrow{\mathrm{F}} \cdot d \vec{r}=\int\left(\overrightarrow{\mathrm{F}}, \frac{d \vec{r}}{d t}\right) d t=\int(\overrightarrow{\mathrm{F}} \cdot \vec{v}) d t\left[v=\frac{d \vec{r}}{d t}\right] \ldots \ldots\) (3)
Substituting equation (2) and equation (3) in equation (1), we get
\(\int \frac{d W}{d t} d t=\int(\overrightarrow{\mathrm{F}} \cdot \vec{v}) d t\)
Or
\(\int\left(\frac{d W}{d t}-\overrightarrow{\mathrm{F}} \cdot \vec{v}\right) d t=0\)
This relation is true for any arbitrary value of dt. This implies that the term within the bracket must be equal to zero, i.e.,
\(\frac{d W}{d t}-\overrightarrow{\mathrm{F}} \cdot \vec{v} =0  \)
\(\frac{d W}{d t} =\overrightarrow{\mathbf{F}} \vec{v}\)
Examples: Motors, Engines and Automobiles
A vehicle of mass 1250 kg is driven with an acceleration 0.2 ms^-2 along a straight level road against an external resistive force 500 N. Calculate the power delivered by the vehicle's engine if the velocity of the vehicle is 30 ms^-1.
Solution
The vehicle's engine has to do work against resistive force and make vehicle to move with an acceleration. Therefore, power delivered by the vehicle engine is
\(P =(\text { resistive force }+\text { mass } \times \text { acceleration }) \text { (velocity) }  \)
\(P =\overrightarrow{\mathbf{F}}_{-\mathrm{ma}} \vec{v}=\left(F_{\text {reistie }}+F\right) \vec{v}  \)
\(P =\overrightarrow{\mathbf{F}}_{\text {tot }} \vec{v}=\left(F_{\text {reithie }}+m a\right) \vec{v}  \)
\(=500 \mathrm{~N}+\left((1250 \mathrm{~kg}) \times\left(0.2 \mathrm{~ms}^{-2}\right)\right)\left(30 \mathrm{~ms}^{-1}\right)=22.5 \mathrm{kw}\)

(i) Linear momentum is conserved in all dollision processes. When two bodies . collide, the mutual impulsive forces acting between them during the collision time (\(\Delta\)t) produces a change in their respective momenta.
(ii) That is, the first body exerts a force \(\overline { { F }_{ 12 } } \) on the second body. From Newton's third law, the second body exerts a force \(\overline { { F }_{ 21 } } \) on the first body. This caus~s a, change in momentum \(\Delta \overrightarrow { { p }_{ 1 } } \) and \(\Delta \overrightarrow { { p }_{ 2 } } \) of the first body and second body respectively. Now, the relations could be written as,
\(\Delta \overrightarrow { { p }_{ 1 } } \) = \(\overline { { F }_{ 12 } } \) \(\Delta\)t  ......(1)
\(\Delta \overrightarrow { { p }_{ 2 } } \) = \(\overline { { F }_{ 21 } } \) \(\Delta\)t  ......(2)
(iii) Adding equation (1) and equation (2), we get
\(\Delta \overrightarrow { { p }_{ 1 } } \) + \(\Delta \overrightarrow { { p }_{ 2 } } \) = \(\overline { { F }_{ 12 } } \) \(\Delta\)t + \(\overline { { F }_{ 21 } } \) \(\Delta\)t = \(\left( \overrightarrow { { F }_{ 12 } } +\overrightarrow { { F }_{ 21 } } \right) \Delta t\)
According to Newton's third law \(\overline { { F }_{ 12 } } \) = - \(\overline { { F }_{ 21 } } \)
\(\Delta \overrightarrow { { p }_{ 1 } } \) + \(\Delta \overrightarrow { { p }_{ 2 } } \) = 0
\(\Delta( \overrightarrow { { p }_{ 1 } } + \overrightarrow { { p }_{ 2 } } )\) = 0
(iv) Dividing both sides by \(\Delta\)t and taking limit \(\Delta\)t \(\rightarrow\) 0 we get
\(\lim _{ \Delta t\rightarrow 0 }{ \frac { \Delta \left( \overrightarrow { { p }_{ 1 } } +\overrightarrow { { p }_{ 2 } } \right) }{ \Delta t } } =\frac { \Delta \left( \overrightarrow { { p }_{ 1 } } +\overrightarrow { { p }_{ 2 } } \right) }{ dt } =0\)
 \(\therefore\)  p₁ + p₂ = constant
(v) The above expression implies that the total linear momentum is a conserved quantity.

(i) A compressed or extended spring will transfer its stored potential energy into, kinetic energy of the mass attached to the spring.
(ii) The potential energy-displacement graph is shown in Figure.

Potential energy-displacement graph for a spring-mass system
(iii) In a frictionless environment, the energy gets transferred from kinetic to potential and potential to kinetic repeatedly such that the total energy of the system remains constant. At the mean position,  ΔKE = ΔU

Applied force = 10 N
Opposing friction forcej= M_k. N = M_k·mg.
= 0.3\(\times\)3\(\times\)9.8 = 8.82 N.
Net accelerating forceF - f = 10N - 8.82N
=1.18N
Acceleration a =\(\frac{force}{mass}\)=\(\frac{8.82N}{3Kg}=2.94 ms^{-2}\)
Distance covered in 10s (assuming w = 0)
\(s=0+\frac{1}{2}at^{2}=\frac{1}{2}\times2.94\times(10^{2}) = 147 m\)
there force workdone by a applied force,
W = Fs = 10\(\times\)147
W = 1470 J

(i) 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.
(ii) Consider a body of mass m being moved from ground to the height h against the gravitational force as shown in Figure.
            
iii) The gravitational force \({ \overrightarrow { F } }_{ g }\) acting on the body is, \({ \overrightarrow { F } }_{ g }=-mg\hat { j } \) (as the force is' in y-direction, unit vector is used). Here, negative sign implies that the force is acting vertically downwards. In order to move the body without acceleration (or with constant velocity), an external applied force \({ \overrightarrow { F } }_{ a }\) equal in magnitude but opposite to that of gravitational force \({ \overrightarrow { F } }_{ a }-{ \overrightarrow { F } }_{ g }\) . This implies that \({ \overrightarrow { F } }_{ a }=-mg\hat { j } \)
(iv) The positive sign implies that the applied force is in vertically upward direction. Hence, when the body is lifted up its velocity remains unchanged and thus its kinetic energy also remains constant.
(v) The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from the ground to that height h.
\(U=\int \overrightarrow { { F }_{ a } } .d\overrightarrow { r } =\int \left| \overrightarrow { { F }_{ a } } \right| \left| d\overrightarrow { r } \right| \cos { \theta } \)
(vi) Since the displacement and the applied force are in the same upward direction, the angle between them, \(\theta ={ 0 }^{ o }\)
Hence, \(\cos { { \theta }^{ o } } =1\) and \(\left| \overrightarrow { { F }_{ a } } \right| =mg\) and
\(=\left| d\overrightarrow { r } \right| dr.\)
\(U=\overset { h }{ \underset { 0 }{ \int } } dr\)
\(U=mg{ \left[ r \right] }_{ 0 }^{ h }\) = mgh.

If there is a loss of kinetic energy during a collision, then it is called as an inelastic collision
In the case of inelastic collision,
(i) Total kinetic energy is not conserved.
(ii) Some or all of the forces involved are non-conservative.
(iii) A part of the mechanical energy is transformed into heat, sound, light etc.
Examples for inelastic collision:
(i) Collision between ball and floor
(ii) Collision between two vehicles
Examples for perfectly inelastic collision:
(i) Mud thrown on a wall and sticking to it
(ii) a man jumping into a moving trolley
(iii) a bullet fired into a wooden block and remaining embedded in it.