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

Question

A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is: (A) 16 J (B) 8 J (C) 32 J (D) 24 J

Tardigrade Admin · Accepted Answer

The work is stored as the P.E. of the body and is given by, U=∫x1x2Fexternaldx Or U=∫x1x2kx dx =(1/2)k(x22-x12) =(800/2)[(0.15)2-(0.05)2] [k=800(given)] =400[0.2× 0.1] =8 joule

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

16 J

8 J

32 J

24 J

The work is stored as the P.E. of the body and is given by, $U = \int_{x_{1}}^{x_{2}} F_{e x t er na l} d x$ Or $U = \int_{x_{1}}^{x_{2}} k x d x$ $= \frac{1}{2} k (x_{2}^{2} - x_{1}^{2})$ $= \frac{800}{2} [(0.15)^{2} - (0.05)^{2}]$ $[k = 800 (g i v e n)]$ $= 400 [0.2 \times 0.1]$ $= 8 j o u l e$

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

16 J

8 J

32 J

24 J

The work is stored as the P.E. of the body and is given by, U=∫x1​x2​​Fexternal​dx Or U=∫x1​x2​​kxdx =21​k(x22​−x12​) =2800​[(0.15)2−(0.05)2] [k=800(given)] =400[0.2×0.1] =8joule

The work is stored as the P.E. of the body and is given by, $U = \int_{x_{1}}^{x_{2}} F_{e x t er na l} d x$ Or $U = \int_{x_{1}}^{x_{2}} k x d x$ $= \frac{1}{2} k (x_{2}^{2} - x_{1}^{2})$ $= \frac{800}{2} [(0.15)^{2} - (0.05)^{2}]$ $[k = 800 (g i v e n)]$ $= 400 [0.2 \times 0.1]$ $= 8 j o u l e$