Question

If the potential energy of a spring when stretched through a distance ‘a’ is 25 J, then what is theamount of work done on the same spring so as to stretch it by an additional distance ‘5a’?

Answer 1

900 J.
Let the stiffness constant of the spring be k
Therefore, the elastic potential energy stored by the spring when it is stretched through a distance of 'a' from its natural length = (1/2) X k X a X a = 25 J (given).
So, k X a X a = 50 J
Now, the total elastic potential energy stored by the spring when it is stretched through an additional distance of '5a' that is the total distance of '6a' from its natural length = (1/2) X k X 6a X 6a  
                                                            =  18 X k X a X a
                                                            =  18 X 50 J
                                                            =  900 J 
 

Answer 2

Answer:

The correct answer is $900J.$

Explanation:

Let the stiffness constant of the spring exist $k$. Therefore, the elastic potential energy accumulated by the spring when it exists stretched through a distance of 'a' from its natural length (given).

$=(1 / 2) \times k \times a \times a=25 \mathrm{~J}$  

So, $k \times a \times a=50 \mathrm{~J}$

Now, the entire elastic potential energy accumulated by the spring when it exists stretched through an extra distance of '5a' that exists the entire distance $' 6 a^{\prime}$ from its natural length

$=(1 / 2) \times k \times 6 a \times 6 a$

$=18 \times k \times a \times a$

$=18 \times 50 J$

$=900 J$

Therefore, the amount of work done on the same spring so as to stretch it by an additional distance ‘5a’ is $900J.$

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