Question

Questions 1 to 4 are based on Table A. Study this table and answer the following questions.

Distance (m)	Height above the base of the mountain (m)	Uniform speed (m/s)
0-500	100	2
500-2000	250	3
2000-4000	450	1.5
4000-5000	500	0.5

Alok is travelling to Vaishnodevi on foot. He starts from the base of the mountain and the temple is at a distance of 5 Km from the base and at a vertical height of 500 m. He also notes his uniform speed, distance and height from the base at regular intervals (shown in table). Alok weighs 50 kg.
1. Find the kinetic energy in the 500-2000 interval.
2. Find his potential energy at the end of 2000-4000 intervals.
3. How much work Alok did against gravity when he reached the summit.
4. State the law of conservation of energy.

Answer 1

Hint:Here, we will be using the formulas for the kinetic energy of a moving body of mass m and moving velocity v as,
${\rm{Kinetic}}\;{\rm{energy}}\;{\rm{ = }}\;\dfrac{1}{2}m{v^2}$ and the potential energy can be calculated by using the formula${\rm{Potential}}\;{\rm{energy}}\;{\rm{ = }}\;{\rm{mgh}}$.

Complete step by step answer:
Given:
We know that speed in 500-2000 interval is given as $v = 3\;{{\rm{m}} {\left/{ {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}$. The weight of the Alok is 50 kg. The height in the interval of 2000-4000 is $h = 450\;{\rm{m}}$ and also the height of the summit is$H = 450\;{\rm{m}}$. The gravitational acceleration is $g = 9.8\;{{\rm{m}} {\left/{ {{\rm{m}} {{\rm{se}}{{\rm{c}}^{\rm{2}}}}}} \right.} {{\rm{se}}{{\rm{c}}^{\rm{2}}}}}$.
$1)$ We can calculate the kinetic energy in the interval 500-2000 as,
$
{\rm{Kinetic}}\;{\rm{energy}}\;{\rm{ = }}\;\dfrac{1}{2}m{v^2}\\
\Rightarrow{\rm{Kinetic}}\;{\rm{energy}}\;{\rm{ = }}\; \dfrac{1}{2}\left( {50} \right){\left( 3 \right)^2}\\
\Rightarrow{\rm{Kinetic}}\;{\rm{energy}}\;{\rm{ = }}\; \dfrac{1}{2}\left( {450} \right)\\
\therefore{\rm{Kinetic}}\;{\rm{energy}}\;{\rm{ = }}\; 225\;{\rm{J}}
$
So, the kinetic energy in the interval 500-2000 is 225 Joule.
$2)$ We can calculate the potential energy at the end of 2000-4000 intervals as,
$
{\rm{Potential}}\;{\rm{energy}}\;{\rm{ = }}\;{\rm{mgh}}\\
\Rightarrow{\rm{Potential}}\;{\rm{energy}}\;{\rm{= 50}}\left( {9.8} \right)\left( {450} \right)\\
\therefore{\rm{Potential}}\;{\rm{energy}}\;{\rm{= 220500}}\;{\rm{J}}
$
So, the potential energy at the end of 2000-4000 intervals is 220500 joules.
$3)$ We can calculate the work done against the gravity when Alok reached the summit as,
$
W = mgH\\
\Rightarrow W = \left( {50} \right)\left( {9.8} \right)\left( {500} \right)\\
\therefore W = 245000\;{\rm{J}}
$
So, the work done against gravity when Alok reached the summit is 245000 joules.
$4)$ According to the law of conservation of energy, energy can only be transformed from one form to another form. It cannot be created or destroyed. In other words the sum of the kinetic energy and potential energy is always conserved if there is no any resistive or friction force acting on the system.

Note: In these types of questions one thing should be remembered that the kinetic energy calculation and potential energy calculations are dependent on the values of mass of the body, linear velocity and datum head of the body from the ground.