Here m = 20 g = 0.02 kg, u = 150 m s^-1, v = 0 and s = 10 cm = 0.1 m
According to work-kinetic energy theorem, we have
K-K' = W = Fs
\(\therefore \quad \frac { 1 }{ 2 } { mu }^{ 2 }-0=Fs\Rightarrow F=\frac { { mu }^{ 2 } }{ 2s } =\frac { 0.02\times { (150) }^{ 2 } }{ 2\times 0.1 } =2250\ N.\)

The kinetic energy of the body will increase when work is done on it in accordance with the work-energy theorem.

K.E. becomes four times.

Collision in which the colliding particles move along the same straight line path before as well as after the collision, is called headon collision.

The objects mutually exchange their velocities as a consequence of collision.

Work done against the gravitational force.

Because centripetal force is perpendicular to the displacement of the body.

Energy stored in a spring is \(\frac { 1 }{ 2 } { kx }^{ 2 }\)

Negative sign indicates that the work is done against the force.

When the external force is removed after the push, the mass gets detached when spring obtain its natural length and say, mass m rises h height from the pushed position.
      
Loss in potential energy of spring = Gain in gravitational potential energy
\(\frac { 1 }{ 2 } k[{ x }^{ 2 }]\ =mgh\)
\(\frac { 1 }{ 2 } \times 200[{ 0.1 }^{ 2 }]\quad =0.5\times 10\times h\)
1 = 5 h \(\Rightarrow \)  h = 0.2 m

In present problem, there are no net external forces. the sand is slowly leaking downwards  from the whole. Consequently, normal  reaction acting vertically upwards also slowly decreases.
But it does not do any work because it is at right angle to trolley's motion. Thus, there is no change in speed of the trolley and it will continue among moving with a  uniform speed of 27 kmh^-1

0.12 J

We can calculate the spring constant of spring, as it ios extended by 20 cm under 60 N force.
F = kx \(\Rightarrow\)60 = k\(\times\) 20 \(\times\)10^-2 
k = 300 N/m
At a force of 20 N, the extension in spring is 
F = kx \(\Rightarrow\) 20 = 300x
x = \(\frac { 2 }{ 30 } =\frac { 1 }{ 15 } m\)

Yes, when p = 0
Then K = \(\frac { { p }^{ 2 } }{ 2m } \\\)=0
But E = K + U = U (potential energy) which may or may not be zero.

\(\left.\begin{array}{l}\text { K.E. of the } \\ \text { bob at its } \\ \text { mean position }\end{array}\right\}=\left\{\begin{array}{l}\text { P.E. at } \\ \text { its highest } \\ \text { position }\end{array}\right.\)
i.e., \(\frac{1}{2} m v^2=m g h\)
Velocity of the bob
\(v  =\sqrt{2 g h} \)
\(=\sqrt{2 \times 9.8 \times 10} \)
\(=\sqrt{1.96} \)
\(v  =1.4 \mathrm{~m} / \mathrm{s}\)