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.

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