A spring of force constant K is first stretched by distance a from its natural length and then further by distance b. The work done in stretching the part b is

Question

A spring of force constant K is first stretched by distance a from its natural length and then further by distance b. The work done in stretching the part b is (A) (1/2) K a(a-b) (B) (1/2) Ka (a+ b) (C) (1/2) K b(a-b) (D) (1/2) K b(2 a +b)

Tardigrade Admin · Accepted Answer

W1=(1/2) k x2=(1/2) k a2 W2=(1/2) k(a+b)2 Δ W=W2-W1=(1/2) k b(2 a+b)

Q. A spring of force constant $K$ is first stretched by distance a from its natural length and then further by distance $b$ . The work done in stretching the part $b$ is

$\frac{1}{2} K a (a - b)$

$\frac{1}{2} K a (a + b)$

$\frac{1}{2} K b (a - b)$

$\frac{1}{2} K b (2 a + b)$

$W_{1} = \frac{1}{2} k x^{2} = \frac{1}{2} k a^{2}$
$W_{2} = \frac{1}{2} k (a + b)^{2}$
$Δ W = W_{2} - W_{1} = \frac{1}{2} kb (2 a + b)$

Q. A spring of force constant K is first stretched by distance a from its natural length and then further by distance b. The work done in stretching the part b is

21​Ka(a−b)

21​Ka(a+b)

21​Kb(a−b)

21​Kb(2a+b)