m = 2 kg, d = 10 m, F_ext = 20 N, \(\mu\)_k = 0.9.
when an object is in motion on he horizontal surface, it experiences two forces.
(a) External force, F_ext = 20 N
(b) Kinetic friction,
f_k = \(\mu\)_k mg = 0.9 \(\times\) (2) \(\times\) 10 = 18N
The work done by the external force W_ext = Fd = 20 x 10 = 200J
The work done by the force of kinetic friction W_k = f_kd = (-18) \(\times\) 10 = -180 J. Here the negative sign implies that the force of kinetic friction is opposite to the direction of displacement.
The total work done on the object W_total = W_ext + W_k = 200 J - 180 J = 20 J.
Since the friction is a non-conservative force, out of 200 J given by the external force, the 180 J is lost and it can not be recovered.

a) When the weight lifter lifts the mass, force and displacement are in the same direction, which means that the angle between them θ = 0°. Therefore, the work done by the weight lifter,
W_{weight lifter} = F_wh cos θ = F_wh (cos 0°)
= 5000\(\times\)5\(\times\)(1) = 25,000 joule= 25 kJ
(b) When the weight lifter lifts the mass, the gravity acts downwards which means that the force and displacement are in opposite direction. Therefore, the angle between them θ = 180°.
W_gravity = F_gh cos θ = mgh( cos 180°)
= 250\(\times\)10\(\times\)5\(\times\)(-1) = -12,500 joule = -12.5 kJ
(c) The net work done (or total work done) on the object
W_net = W_{weight lifter}+ W_gravity
= 25 kJ -12.5 kJ = +12.5 kJ

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.

(a) Since the spring force is a conservative force, the total energy is constant. At x = 10 m, the total energy is purely potential.
\(E=U=\frac { 1 }{ 2 } { kx }^{ 2 }=\frac { 1 }{ 2 } \times (1)\times ({ 10 })^{ 2 }=50\quad J\)
When the mass crosses the equilibrium position (x = 0) the potential energy
\(U=\frac { 1 }{ 2 } \times 1\times (0)=0\quad J\)
The entire energy is purely kinetic energy at this position.
\(E=KE=\frac { 1 }{ 2 } { mv }^{ 2 }=50\quad J\)
The speed \(v=\sqrt { \frac { 2KE }{ m } } =\sqrt { \frac { 2\times 50 }{ 2 } } =\sqrt { 50 } { ms }^{ -1 }=7.07 \ { ms }^{ -1 }\)
(b) Since the restoring spring force is F = -kx, when the object crosses the equilibrium position, it experiences no force. Note that at equilibrium position, the object moves very fast. When the object is at x = +10m (elongation), the force F = -kx.
F = -(1)(10) =-10 N. Here the negative sign implies that the force is towards equilibrium
i.e., towards negative x-axis and when the object is at x = -10 (compression), it experiences a forces F = -(1) (-10) = +10 N. Here the positive sign implies that the force points towards positive x-axis.
The object comes to momentary rest at x = ±10 m even though it experiences a maximum force at both these points.

m = 100 kg, h = 10 m
Along path (1):
The minimum force F₁ required to move the object to the height of 10 m should be equal to the gravitational force, F₁ = mg = 100 \(\times\) 10 = 1000 N.
The distance moved along path (1) is, h = 10 m.
The work done on the object along path (1) is
W = Fh = 1000 \(\times\) 10 = 10,000 J
Along path (2):
In the case of the ramp, the minimum force F₂ that we apply on the object to take it up is not equal to mg, it is rather equal to mg sin \(\theta\). (mg sin \(\theta\) < mg).
Here, angle \(\theta\) = 30°
Therefore, F₂ = mg sin \(\theta\) = 100 \(\times\)10 \(\times\) sin30° = 100 \(\times\) 10 \(\times\) 0.5 = 500 N.
Hence, (mg sin \(\theta\) < mg)
The path covered along the ramp is, \(l=\frac { h }{ sin \ 30^o } =\frac { 10 }{ 0.5 } =20 \ m\)
The work done on the object along path (2) is, W = F₂ l = 500 \(\times\) 20 = 10,000 J.
Since the gravitational force is a conservative force, the work done by gravity on the object is independent of the path taken.
In both the paths the work done by the gravitational force is 10,000 J.
Along path (1): more force needs to be applied against gravity to cover lesser distance.
Along path (2): lesser force needs to be applied against the gravity to cover more distance,
As the force needs to be applied along the ramp is less, it is easier to move the object along the ramp.

The spring constant, k = 0.1 N m^-1
The displacement, x = 25 cm = 0.25 m
(a) The potential energy stored in the spring is given by
\(U=\frac { 1 }{ 2 } { kx }^{ 2 }=\frac { 1 }{ 2 } \times 0.1\times { (0.25) }^{ 2 }=0.0031J\)
(b) The work done W_s by the spring force _{\(\bar { F } \)} is given by,
\({ W }_{ s }=\int _{ 0 }^{ x }{ \overrightarrow { { F }_{ s } } .\overrightarrow { dr } } =\int _{ 0 }^{ x }{ (-k\ x\hat { i } ).(dx\hat { i } ) } \)
The spring force \(\overrightarrow { { F }_{ s } } \)  acts in the negative x direction while elongation acts in the positive x direction.
\({ W }_{ s }=\int _{ 0 }^{ x }{ (-kx)dx=-\frac { 1 }{ 2 } { kx }^{ 2 } } \)
\({ W }_{ s }=-\frac { 1 }{ 2 } \times 0.1\times { (0.25) }^{ 2 }=-0.0031\ J\)
Note that the potential energy is defined through the work done by the external agency. The positive sign in the potential energy implies that the energy is transferred from the agency to the object. But the work done by the restoring force in this case is negative since restoring force is in the opposite direction to the displacement direction.
(c) During compression also the potential energy stored in the object is the same.
\(U=\frac { 1 }{ 2 } { kx }^{ 2 }=0.0031\ J\)
Work done by the restoring spring force during compression is given by
\({ W }_{ s }=\int _{ 0 }^{ x }{ { \overrightarrow { F } }_{ s } } \overrightarrow { dr } =\int _{ 0 }^{ x }{ (kx\hat { i } ).(-dx\hat { i } ) } \)
In the case of compression, the restoring spring force acts towards positive x-axis and displacement is along negative x direction.
\({ W }_{ s }=\int _{ 0 }^{ x }{ (-kx)dx=-\frac { 1 }{ 2 } { kx }^{ 2 } } =-0.0031J\)

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.
\(\overrightarrow { { F }_{ g } } +\overrightarrow { { F }_{ a } } =0\)

(a) The kinetic energy of the mass is given by \(KE=\frac { { p }^{ 2 } }{ 2m } \)
For the object of mass 2 kg, kinetic energy is KE₁ = \(\frac { ({ 20 })^{ 2 } }{ 2\times 2 } =\frac { 400 }{ 4 } =100 \ J\)
For the object of mass 4 kg, kinetic energy is KE₂ = \(\frac { { (20) }^{ 2 } }{ 2\times 4 } =\frac { 400 }{ 8 } =50 \ J\)
Note that KE₁ \(\neq \) KE₂ i.e., even though both are having the same momentum, the kinetic energy of both masses is not the same. The kinetic energy of the heavier object has lesser kinetic energy than smaller mass. It is because the kinetic energy is inversely proportional to the mass (KE \(\infty \frac { 1 }{ m } \) ) for a given momentum.
(b) As the momentum, p = mv, the two objects will not have same speed.

Force \(\overrightarrow { F } =mg\left( -\hat { j } \right) =-mg\hat { j } \)
Displacement vector \(d\vec{r}\) = dx\(\hat { i } \) + dy \(\hat { j } \)
(As the displacement is in two dimension; unit vectors \(\hat { j } \) and \(\hat { i } \)  are used)
(a) Since the motion is only vertical, horizontal displacement component dx s zero. Hence, work done by the force along path 1 (of distance h).
\({ W }_{ push\ \ 1 }=\int _{ A }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ B }{ (-mg\hat { j } ).(dy\hat{j})=-mg\int _{ 0 }^{ h }{ dy=-mgh } } } \)
Total work done for path 2 is
\({ W }_{ push\ \ 2 }=\int _{ A }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ C }{ \overrightarrow { F } .d\overrightarrow { r } +\int _{ C }^{ D }{ \overrightarrow { F } .d\overrightarrow { r } +\int _{ D }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } } } } } \)
But \(\int _{ A }^{ C }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ B }{ (-mg\hat { j } ).(dx\hat { i } )=0 } } \)
\(\int _{ A }^{ D }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ C }^{ D }{ (-mg\hat { j } ).(dy\hat { j } )=-mg\int _{ 0 }^{ h }{ dy } =-mgh } } \)
\(\int _{ D }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ B }{ (-mg\hat { j } ).(dx\hat { i } )=0 } } \)
Therefore, the total work done by the force along the path 2 is
\({ W }_{ push\ \ 2 }=\int _{ A }^{ B }{ \vec { F } } .d\overrightarrow { r } =-mgh\)
Note that the work done by the conservative force is independent of the path.