Question

Questions 14.1 to 14.4 are based on the Table A.
Study this table and answer the following questions:
Table A
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.

14.1 Find the kinetic energy in the 500 – 2000interval.
[1]
14.2 Find his potential energy at the end of 2000-4000interval.
[1]
14.3 How much work has Alok done against the gravity when he reaches the summit?

14.4 State the law of conservation of energy.​

Answer 1

Answer:

1) Kinetic energy in interval 500 - 2000 is 225 J

2) Potential energy in interval 2000 - 4000 is 220500 J

3)Work done against gravity in reaching the summit is 245000 J

Explanation:

1) Kinetic energy is given by

$KE = \frac{1}{2}mv^2$

As we know that speed in 500 - 2000 interval is given as

v = 3 m/s

so we have

$KE = \frac{1}{2}(50)(3^2)$

$KE = 225 J$

2)Potential energy is given as

$PE = mgH$

As we know that height in the interval of 2000 - 4000 interval is

H = 450 m

$PE = (50)(9.8)(450)$

$PE = 220500 J$

3) Work done against gravity is given as

$W = mgH$

Height of summit is given as

H = 500 m

$W = (50)(9.8)(500)$

$W = 245000 J$

4) As per law of conservation of energy sum of potential energy and kinetic energy is always conserved if there is no resistive force or there is no friction force on it.

#Learn

Topic : Work Energy Power

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