First slide

Work done by Diff type of forces

Question

A spring of spring constant 5x10³ N/m is stretched initially by 5 cm from the unscratched position. Then the work required to stretch it further by another 5 cm is

Moderate

A

$6 â 25 N â m$
B

$12 â 50 N â m$
C

18 . 75 N-m
D

25 '00 N-m

Solution

for a spring .U $= \frac{1}{2} {kx}_{1}^{2}$
where x1 =5 cm
When the spring is further stretched by 5 cm, then
$U^{â²} = \frac{1}{2} {kx}_{2}^{2}$
Now $W = U^{â²} â U = \frac{1}{2} k (x_{2}^{2} â x_{1}^{2})$
$= \frac{1}{2} (5 Ã 10^{3}) [{(10 Ã 10^{â 2})}^{2} â {(5 Ã 10^{â 2})}^{2}] = 18.75 N m$

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