Activity 11.15 Class 9 Science Chapter 11 Work and Energy

Activity 11.15 Class 9 Science Chapter 11

Activity 11.15 Class 9 Science belongs to Chapter 11 Work and Energy Class 9 Science. You will get a complete solution with an explanation and conclusion. After reading activity 11.15, you will be able to answer the questions based on the activity.

Activity 11.15 Class 9 Science Work and Energy

 

You are suggested to study NCERT activity 11.15 class 9 science chapter 11Work and Energy explanation with a conclusion so that you can attempt questions based on activity 11.15 class 9 science.

Activity 11.15 Class 9 Science

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

      

   ·     For simplifying the calculations, take the value of g as 10 m/s².

Calculation and explanation of Activity 11.15 Class 9 Science

`E_p=mgh` 

`E_k=1/2 mv^2` 

Now we can calculate for each situation given in the table above

(a).  When h= 4m

`E_p=mgh` = 20× 10 ×4 

`E_p=800J` 

`E_k=0` 

(b) When h=3m

`E_p=mgh` = 20× 10 ×3   

`E_p=600J` 

`E_k=1/2 mv^2`= mgs 

`E_k=20×10×1=200J` (here s=  4m-3m=1m)

(c ) When h= 2m

`E_p=mgh` = 20× 10 ×2   

`E_p=400J` 

`E_k=1/2 mv^2`= mgs 

`E_k=20×10×2=400J` (here s=  4m-2m=2m)

 (d) When h=  1m

`E_p=mgh` = 20× 10 ×1   

`E_p=200J` 

`E_k=1/2 mv^2`= mgs 

`E_k=20×10×3=600J` (here s=  4m-1m=3m)

(e) When h=0

`E_p=mgh` = 20× 10 ×0   

`E_p=0` 

`E_k=1/2 mv^2`= mgs 

`E_k=20×10×4=800J` (here s=  4m)

Height at which the object is situated (m)	Potential energy `E_p=mgh` (J)	Kinetic energy  `E_k=1/2 mv^2`  (J)	`E_p+E_k`   (J)
4	800	0	800
3	600	200	800
2	400	400	800
1	200	600	800
0	0	800	800

Conclusion of activity 11.15 class 9 science

This activity shows that when energy is transformed, the total energy remains unchanged.

Related Topics 

1. Activity 11.16 Class 9 Science

2. Activity 11.14 Class 9 Science

2. NCERT Solutions of chapter 11