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 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.

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 (resistive force + mass x acceleration) (velocity)
\(P=\overrightarrow { { F }_{ -tot } } \overrightarrow { v } =\left( { F }_{ resistance }+F \right) \overrightarrow { v } \)
\(P=\overrightarrow { { F }_{ tot } } .\overrightarrow { v } =\left( { F }_{ resistance }+ma \right) \overrightarrow { v } \)
= (500 N + (1250 kg) \(\times\) (0.2 ms^-2)) (30 ms^-1) = 22.5 kW