1.

A spring of force constant 800 N/m has an extension of 5 cm. The workdone in extending it from 5 cm to 15 cm is :

Answer»

16 J
8 J
32 J
24 J

Solution :Here, the WORK done is streching a spring is stored as the potential ENERGY of spring i.e `1/2kx^2`
Thus, `W=1/2kx_2^2-1/2kx_1^2`
= `1/2k(x_2^2-x_1^2)`
=`1/2xx800[(15/100)^2-(5/100)^2]`
=`1/2xx(800)/(10^4)[225-25]`
=`4XX10^(-2)xx200=8 J`


Discussion

No Comment Found

Related InterviewSolutions