Solution to Activity 10.15 of NCERT Class 9 Science Chapter 10 Work and Energy

15. Complete Activity 10.15 (Page 122).

An object of mass 20 kg is dropped from a height of 4 m. Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case.

Height at which object is located m	Potential energy (E_p = mgh) J	Kinetic energy (E_k = mv²/2) J	E_p + E_k J
4
3
2
1
Just above the ground

Answer:

Height at which object is located m	Potential energy (E_p = mgh) J	Kinetic energy (E_k = mv²/2) J	E_p + E_k J
4	800	0	800
3	600	200	800
2	400	400	800
1	200	600	800
Just above the ground	0	800	800

Answer:

At 4 m:

E_p = mgh = 20 × 10 × 4 = 800 J.

v = 0

E_k = mv²/2 = 0

E_p + E_k = 800 J + 0 = 800 J.

At 3 m:

E_p = mgh = 20 × 10 × 3 = 600 J.

v² = u² + 2gs

v² = 0² + 2 × 10 × 1

v² = 20

v = 2√5 m/s

E_k = mv²/2 = 20 × 20/2 = 200 J

E_p + E_k = 600 J + 200 J = 800 J

At 2 m:

E_p = mgh = 20 × 10 × 2 = 400 J.

v² = u² + 2gs

v² = 0² + 2 × 10 × 2

v² = 40

v = 2√10 m/s

E_k = mv²/2 = 20 × 40/2 = 400 J

E_p + E_k = 400 J + 400 J = 800 J

At 1 m:

E_p = mgh = 20 × 10 × 1 = 200 J.

v² = u² + 2gs

v² = 0² + 2 × 10 × 3

v² = 60

E_k = mv²/2 = 20 × 60/2 = 600 J

E_p + E_k = 200 J + 600 J = 800 J

Just above the ground:

E_p = mgh = 20 × 10 × 0 = 0 J.

v² = u² + 2gs

v² = 0² + 2 × 10 × 4

v² = 80

E_k = mv²/2 = 20 × 80/2 = 800 J

E_p + E_k = 0 J + 800 J = 800 J

“An object of mass 20 kg is dropped from a height of 4 m. Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case.” – Solved.