As the energy is recoverable therefore the force is conservative force.

Kinetic energy of a body cannot be negative because it is given by \(\frac { 1 }{ 2 } \) mv² and is always positive whether v is -ve or +ve. The gravitational potential energy of a body may be negative or positive.

Here, force F = 1500N
and  Displacement   , s = 100 m, \(\theta\) = 60⁰
\(\therefore\) Work done , W = Fs cos \(\theta\)
= 1500\(\times\)100\(\times\)cos(60⁰)
= 1500\(\times\)100\(\times\) \(\frac{1}{2} \)
= 75000 J
= 75 kJ

The winning team ( i.e the team which is pulling the other team towards itself ) is doing positive work and the losing team ( i.e. the team slowly giving way to the other ) is doing negative work. )

Here, the volume of water raised V = 2000 L
Density of water, p =1kg/L
Therefore mass of water raised, m = vp = 2000 x 1 = 2000kg
Power, \(P=\frac { W }{ t } =\frac { mgh }{ t } =\frac { 2000\times 9.8\times 120 }{ 5\times 60 } =7840 \ W\ \)
\(\ 7.840\ kW\ \ [1kW=1000\quad W\)

During the short interval of an elastic collision, total linear momentum is conserved.

Collision in which momentum and total energy remained onserved and total kinetic energy of the colliding particles remain constant both before and after the collision, is clled elastic collision.

The value is almost always less than one due to initial translational kinetic energy being lost to rotational kinetic energy, plastic deformation, and heat. It can be more than 1 if there is an energy gain during the collision from a chemical reaction, a reduction in rotational energy, or another internal energy decrease that contributes to the post-collision velocity.
\(\text { Coefficient of restitution }(e)=\frac{\mid \text { Relative velocity after collision } \mid}{\mid \text { Relative velocity before collision } \mid}\)

Total energy at B = kinetic energy + potential energy
= 0 + mgh
= 2 \(\times\) 9.8 \(\times\)10
= 196 J
As it descends half the height, it loses potential energy which is given by
\(=mg\frac { h }{ 2 } \)
\(=\frac { 1 }{ 2 } mgh=98\ J\)
\(\therefore\) its potential energy at C = (196 - 98) = 98 J
The loss of potential energy = gain in kinetic energy
= 196 - 98
= 98 J
But \(K.E.=\frac { 1 }{ 2 } { mv }^{ 2 }\)
\(\therefore \quad \frac { 1 }{ 2 } \times 2\times { v }^{ 2 }=98\)
\(\Rightarrow \ { v }^{ 2 }=98\ or\ v=7\sqrt { 2 } m/s\)

We can calculate spring constant of spring by first information.
  
Work done by a vertical spring is = 0.3 g = k[10\(\times \) 10^-2]
2 = k\(\times \)0.1\(\Rightarrow \) k = \(\frac { 3 }{ 0.1 } =30\frac { N }{ m } \)
Extra work required to stretch it by next 5 cm.
W = \(\frac { 1 }{ 2 } k{ x }_{ 2 }^{ 2 }\)-\(\frac { 1 }{ 2 } k{ x }_{ 1 }^{ 2 }\)
W = \(\frac { 1 }{ 2 } \times 30\left[ { \left( 15\times { 10 }^{ -2 } \right) }^{ 2 }-{ \left( 10\times { 10 }^{ -2 } \right) }^{ 2 } \right] \)
W = 15[225-100] \(\times { 10 }^{ -4 }\) = 15\(\times \)125\(\times { 10 }^{ -4 }\) J
W = 0.1875 J

Here, \(m_1=1.2 \mathrm{~kg}, u_1=20 \mathrm{~cm} / \mathrm{s}, m_2=1.2 \mathrm{~kg}, u_2=0\)
If v₁ and v₂ are velocities of the two blocks after collision, then accordin to the principle of conservation of momentum, \(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2\)
\(1.2 \times 20+0=1.2 v_1+1.2 v_2 \therefore v_1+v_2=20(\mathrm{~cm} / \mathrm{s}) \ldots(\mathrm{i})\)
velocity of approach \(=u_1-u_2=20 \mathrm{~cm} / \mathrm{s}\), velocity of separation \(=v_2-v_1\)
By definition, \(e=\frac{v_2-v_1}{u_1-u_2}\)
\(\frac{3}{5}=\frac{v_2-v_1}{20} \therefore v_2-v_1=\frac{20 \times 3}{5}=12 \text {.. }\)
From (i) and (ii), \(v_1=4 \mathrm{~cm} / \mathrm{s}, v_2=16 \mathrm{~cm} / \mathrm{s}\)
Loss in K.E.
\( =\frac{1}{2} m_1 u_1^2-\frac{1}{2}\left(m_1\right) v_1^2-\frac{1}{2} m_2 v_2^2=\frac{1}{2} \)
\( \times 1.2\left(\frac{20}{100}\right)^2-\frac{1}{2}(1.2)\left(\frac{4}{100}\right)^2-\frac{1}{2} \times 1.2\left(\frac{16}{100}\right)^2 \)
\( =2.4 \times 10^{-2}-0.096 \times 10^{-2}-1.536 \times 10^{-2}\)
\(7.7\times { 10 }^{ -3 }\)

At maximum compression the kinetic energy of the car is converted entirely into the potential energy of the spring.
The kinetic energy of the moving car is
\(K=\frac{1}{2} m v^{2}\)
\(=\frac{1}{2} \times 10^{3} \times 5 \times 5\)
K = 1.25 x 10⁴ J
where we have converted 18 km h^–1 to 5 m s^–1 [It is useful to remember that 36 km h^–1 = 10 m s^–1]. At maximum compression x_m , the potential energy V of the spring is equal to the kinetic energy K of the moving car from the principle of conservation of mechanical energy.
\(V=\frac{1}{2} k x_{m}^{2}\)
= 1.25 x 10⁴ J
We obtain
x_m = 2.00 m
We note that we have idealised the situation. The spring is considered to be massless. The surface has been considered to possess negligible friction.

The forces acting on the block are shown in Figure. As the block moves with uniform velocity the forces add up to zero.
\(\therefore \ F\ cos\ \theta =\mu N\)         ...........(i)
F sin \(\theta\) + N = Mg               ...........(ii)
Eliminating N from equtions (i) and (ii)
F cos \(\theta\) = \(\mu\) (Mg - F sin \(\theta\))
\(F=\frac { \mu Mg }{ cos\theta +\mu sin\theta } \)
Work done by this force during a displacement d
\(W=F.d\ cos\ \theta =\frac { \mu Mgd\ cos\theta }{ cos\theta +\mu sin\theta } \)

(i) False, internal as well as external forces can change the kinetic energy.Again forces of conservative may change the potential energy of a system.
(ii) False, for non-conservative force the work done over a closed loop is not zero.
(iii) 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.