The force constant of a wire is k and that of another wire of the same material is 2 k. When both the wires are stretched, then work done is

Question

The force constant of a wire is k and that of another wire of the same material is 2 k. When both the wires are stretched, then work done is (A) W 2=2 W 12 (B) W 2=2 W 1 (C) W 2= W 1 (D) W 2=0.5 W 1

Tardigrade Admin · Accepted Answer

Work done in stretching the wire W=(1/2) × force constant × x2 For first wire, W1=(1/2) × k x2=(1/2) k x2 For second wire, W2=(1/2) × 2 k × x2=k x2 Hence, W2=2 W1

Q. The force constant of a wire is $k$ and that of another wire of the same material is $2 k$ . When both the wires are stretched, then work done is

$W_{2} = 2 W_{1}^{2}$

$W_{2} = 2 W_{1}$

$W_{2} = W_{1}$

$W_{2} = 0.5 W_{1}$

Work done in stretching the wire
$W = \frac{1}{2} \times$ force constant $\times x^{2}$
For first wire, $W_{1} = \frac{1}{2} \times k x^{2} = \frac{1}{2} k x^{2}$
For second wire, $W_{2} = \frac{1}{2} \times 2 k \times x^{2} = k x^{2}$
Hence,
$W_{2} = 2 W_{1}$

Q. The force constant of a wire is k and that of another wire of the same material is 2k. When both the wires are stretched, then work done is

W2​=2W12​

W2​=2W1​

W2​=W1​

W2​=0.5W1​